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In coordinates, write a differential form as $\eta ={\sum }_I{\eta }_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$. Its exterior derivative isFor $r=0$ this is the ordinary differential $df$. The chain rule, together with cancellation of symmetric second derivatives against the exterior product, proves that these formulas agree under coordinate changes. They give an $\mathbb{R}$-linear map $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ with $d^2=0$ and $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^r\alpha \wedge d\beta$ when $\alpha$ has degree $r$.

For a differential one-form $\omega ={\sum }_j{\omega }_{j}d{x}^j$, evaluation givesExpanding $X(\omega (Y))-Y(\omega (X))$ gives these coefficient derivatives plus ${\sum }_j{\omega }_j{\sum }_i(X^i{\partial }_i{Y}^j-Y^i{\partial }_i{X}^j)$. The latter is $\omega ([X,Y])$ by the Lie bracket of vector fields formula. Hence

A connection on a vector bundle is an $\mathbb{R}$-linear map $D:\Gamma (E)\to {\Omega }^1(M;E)$ satisfying $D(f\sigma )=df\otimes \sigma +fD\sigma$. Evaluation on a vector field gives $D_X\sigma$, which is $C^{\infty }$-linear in $X$ and satisfies $D_X(f\sigma )=X(f)\sigma +f{D}_X\sigma$. The covariant exterior derivative extends this to bundle-valued forms byThis respects $(f\alpha )\otimes \sigma =\alpha \otimes (f\sigma )$: applying the rule on either side gives the same extra term $df\wedge \alpha \otimes \sigma$. It therefore defines an operator on bundle-valued forms independently of their local tensor representations. In a local frame with connection matrix $A$, its coefficient-column formula is $d^E=d+A\wedge$.

Define the curvature form of a connection by $R\sigma =(d^E{)}^2\sigma$. The extra terms in $(d^E{)}^2(f\sigma )$ cancel, so this equals $f(d^E{)}^2\sigma$. Equivalently the local formula iswhich acts fibrewise on the section coefficients. Thus curvature is a smooth endomorphism-valued two-form, a section of ${\Lambda }^2{T}^{\ast }M\otimes \mathrm{End}(E)$.

For an $E$-valued one-form $a$, applying the ordinary one-form formula in a frame givesTaking $a=D\sigma$ proves the required curvature identityThe bracket correction is what makes the right side tensorial in the vector fields and the section.

Now define the endomorphism bundle connection by $({\tilde{D}}_X\alpha )(s)=D_X(\alpha s)-\alpha (D_{X}s)$. This expression is $C^{\infty }$-linear in $s$: the two terms involving $X(f)\alpha (s)$ cancel when $s$ is replaced by $fs$. It is therefore a smooth bundle endomorphism. It is $C^{\infty }$-linear in $X$, andThese identities prove it is a connection on a vector bundle. In a local frame it reads ${\tilde{D}}_X\alpha =X(\alpha )+[A(X),\alpha ]$.

For the full commutator identity for an endomorphism connection, expand once more:Subtracting the expression with $X,Y$ interchanged cancels both middle terms. ThusHere $D_X,D_Y$ are differential operators on sections, and the square brackets are their operator commutators. In general they are not fibrewise bundle endomorphisms; the curvature operator becomes fibrewise linear after subtracting $D_{[X,Y]}$.

Subtracting ${\tilde{D}}_{[X,Y]}\alpha$ from this identity gives the curvature of an endomorphism bundle connection:Consequently $\tilde{R}=0$ precisely when every $R_p(X,Y)$ commutes with every endomorphism of $E_p$. Arbitrary such endomorphisms can be extended locally in a frame, so this is a fibrewise assertion. A matrix commuting with all diagonal matrix units is diagonal; commuting also with all off-diagonal matrix units forces all its diagonal entries to coincide. The centre of the full endomorphism algebra therefore consists of scalar matrices. For rank $k>0$, define the smooth two-form $\omega =k^{-1}\mathrm{tr}R$. We obtain the scalar-curvature criterion for a flat endomorphism connectionThe converse follows immediately because scalar endomorphisms commute with all endomorphisms. For the zero bundle both curvatures vanish and one can take $\omega =0$. Scalar curvature here means the scalar endomorphism-valued curvature of this bundle connection, not the scalar trace curvature of a Riemannian manifold.

In coordinates, expanding $d\alpha$ and the Lie bracket of vector fields showsThe terms differentiating components of $X,Y$ cancel against the bracket, leaving $({\partial }_i{\alpha }_j-{\partial }_j{\alpha }_i)X^i{Y}^j$. This is the exterior derivative of a one-form evaluated on vector fields.

A connection on a vector bundle is a linear map $\nabla :\Gamma (E)\to {\Omega }^1(E)$ satisfying $\nabla (fs)=df\otimes s+f\nabla s$; evaluation on $X$ defines ${\nabla }_{X}s$. Its curvature form of a connection is $d_A^2$, locally $F_A=dA+A\wedge A$. Direct expansion, using the displayed exterior-derivative identity, givesThe derivative terms on a scalar multiplying $s$ cancel, and the expression is also linear over smooth functions in $X,Y$. Thus it is an alternating tensor with values in $\mathrm{End}(E)$, namely an element of ${\Omega }^2(\mathrm{End}(E))$.

The dual connection and tensor product connection induce the endomorphism bundle connection on $E^{\ast }\otimes E$:Expanding twice cancels the cross terms and givesHere $\varphi ,s$ are local smooth sections; derivatives are not defined for isolated fiber elements without extensions. Subtract the corresponding ${\tilde{\nabla }}_{[X,Y]}$ identity to get the curvature of an endomorphism bundle connection:This vanishes for all $\varphi$ exactly when each $F_A(X,Y)$ is central in the full matrix algebra, hence scalar. For rank $r>0$, $\omega =r^{-1}\mathrm{tr}{F}_A$ is a smooth two-form andConversely scalar curvature commutes with every endomorphism. This is the scalar-curvature criterion for a flat endomorphism connection; it permits a nonflat connection on $E$ itself.

The endomorphism bundle connection of a rank-$r>0$ bundle is flat if and only if the original curvature satisfies $F_A=\omega \mathrm{id}$. By the curvature of an endomorphism bundle connection, flatness means each $F_A(X,Y)$ commutes with every fiber endomorphism. The centre of the full matrix algebra consists of scalar matrices, and $\omega =r^{-1}\mathrm{tr}{F}_A$ is the required smooth two-form. The original connection can still have nonzero scalar curvature.