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The endomorphism bundle connection is ${\tilde{D}}_X\alpha =D_X\circ \alpha -\alpha \circ D_X$ on sections. Expanding twice makes the two middle terms cancel, proving the displayed commutator identity. Subtracting the derivative along the Lie bracket of vector fields gives $\tilde{R}(X,Y)\alpha =[R(X,Y),\alpha ]$. The operators $D_X$ are differential operators, rather than fibrewise endomorphisms; the corrected curvature operator is fibrewise linear.

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.

For the endomorphism bundle connection, $({\tilde{\nabla }}_X\varphi )(s)={\nabla }_X(\varphi s)-\varphi ({\nabla }_{X}s)$. Expanding two derivatives and subtracting ${\tilde{\nabla }}_{[X,Y]}$ givesThus its curvature form of a connection is the commutator action of the original curvature.

The vector bundle whose fibre at $p$ consists of linear maps $E_p\to E_p$. If a local frame changes by $e^{\prime }=eg$, the coefficient matrix of an endomorphism changes by $T^{\prime }=g^{-1}Tg$, giving smooth fibre-linear transitions. For a line bundle this bundle is canonically the trivial scalar bundle, because every fibre endomorphism is scalar multiplication. A connection on a vector bundle induces the endomorphism bundle connection.

Exterior product of differential forms combined with endomorphism composition is associative and generally noncommutative. For an endomorphism-valued one-form $a$, $(a\wedge a)(u,v)=[a(u),a(v)]$. The endomorphism bundle connection extends as a graded derivation of this algebra.

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.

A connection on a vector bundle is a linear operator $\nabla :\Gamma (E)\to {\Omega }^1(M;E)$ satisfying $\nabla (fs)=df\otimes s+f\nabla s$. Choose coordinates $x^i$ and a local frame $e_1,\dots ,e_m$, and define the connection matrix by $\nabla e_b={\sum }_a{A}^a{}_b\otimes e_a$, with $A^a{}_b={\sum }_i{A}^a{}_{bi}d{x}^i$. For a section $s={\sum }_a{s}^a{e}_a$, the coefficient column satisfiesThe symbol $d$ on the right means the ordinary exterior derivative of the coefficient functions.

Extend uniquely to $E$-valued differential forms by the graded Leibniz ruleIn the same local trivialization, an $E$-valued form is a column $\sigma =({\sigma }^a)$ and the covariant exterior derivative isMoving the one-form $A$ past the scalar $r$-form produces exactly the $(-1{)}^r$ in the graded rule. This explains the sign rather than replacing the rule by an ungraded product formula.

The endomorphism bundle has the induced endomorphism bundle connection characterized on sections by ${\nabla }_X^{\mathrm{End}}(T)s={\nabla }_X(Ts)-T{\nabla }_{X}s$. Its coefficient formula is $dT+AT-TA$. On an endomorphism-valued $r$-form $T$, extend as a graded derivation of the endomorphism-valued exterior product, givingExplicitly its $(a,b)$ entry is $d{T}^a{}_b+{\sum }_c{A}^a{}_c\wedge T^c{}_b-(-1{)}^r{\sum }_c{T}^a{}_c\wedge A^c{}_b$. This version is characterized byThe endomorphism-valued version also satisfies $d_A(T\wedge S)=d_{A}T\wedge S+(-1{)}^{r}T\wedge d_{A}S$. Products here combine composition of endomorphisms with wedge products of coefficient forms; matrix factors must retain their order.

These local formulas are independent of the local frame. For $e^{\prime }=eg$, the change of frame of a vector-bundle connection gives $A^{\prime }=g^{-1}Ag+g^{-1}dg$, ${\sigma }^{\prime }=g^{-1}\sigma$ and $T^{\prime }=g^{-1}Tg$. Using $d(g^{-1})=-g^{-1}(dg)g^{-1}$ gives $d_{A^{\prime }}{\sigma }^{\prime }=g^{-1}{d}_A\sigma$ and $d_{A^{\prime }}{T}^{\prime }=g^{-1}(d_{A}T)g$. Thus both extended operators are globally defined.

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.

The connection difference as an endomorphism-valued one-form follows because the graded Leibniz rule terms cancel:Thus $a=D_1-D$ is a global endomorphism-valued one-form. To define the endomorphism bundle connection, requireThe right side is $C^{\infty }$-linear in $s$, so it is indeed an endomorphism-valued one-form, and replacing $\varphi$ by $f\varphi$ gives $df\otimes \varphi +f\tilde{D}\varphi$. In a frame with connection matrix $A$, its extension to an endomorphism-valued $k$-form $b$ isThe endomorphism-valued exterior product combines the wedge of differential forms with composition in $\mathrm{End}E$. In particular,Expanding $d(A+a)+(A+a)\wedge (A+a)$ proves the curvature difference formula:

The Chern connection is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. Take Hermitian inner products linear in the first variable. In a holomorphic local frame write $⟨eu,ev⟩=v^{†}Hu$, with $H$ a positive Hermitian matrix. The second condition forces $A^{0,1}=0$, and metric compatibility isIts $(1,0)$ component determines $A$ uniquely:Conversely this formula and its conjugate transpose satisfy the metric equation. Under a holomorphic local frame change, $H^{\prime }=g^{†}Hg$ and $\partial g^{†}=0$, so the displayed formula transforms by $A^{\prime }=g^{-1}Ag+g^{-1}\partial g$. Hence the local matrices patch, proving existence and uniqueness.

Differentiating $H^{-1}$ givesTherefore the $(2,0)$ vector-bundle curvature is zero. There is no $(0,2)$ component, andThe last equality is ${\bar{\partial }}^{2}A=0$ in holomorphic local frames, so is intrinsic. This is Dolbeault closedness of Chern curvature. The induced endomorphism bundle connection is itself the Chern connection for the Hilbert-Schmidt metric on $\mathrm{End}E$: the tensor product of a metric-compatible connection and its dual preserves the induced metric, and its $(0,1)$ part is the induced holomorphic Dolbeault operator. These two properties and uniqueness suffice; no further matrix calculation is needed.

For two metrics, both Chern connections have the same $(0,1)$ part, so $a\in A^{1,0}(\mathrm{End}E)$. The $(1,1)$ part of the curvature difference formula is just ${\bar{\partial }}_{\mathrm{End}E}a$; the other two terms have type $(2,0)$, which must cancel because both vector-bundle curvatures have type $(1,1)$. ThusBy Dolbeault theorem, this proves that the curvature Dolbeault classis independent of the metric.

Endomorphism composition and wedge product of differential forms define the powers in $A^{k,k}(\mathrm{End}E)$, rather than in a tensor power of the endomorphism bundle. The Dolbeault operator is a derivation, so $\bar{\partial }({\Theta }^k)=0$. The noncommutative telescoping identity gives the stronger explicit independence statementAll powers of vector-bundle curvature have even total degree, so the derivation introduces no extra signs before $a$. Consequentlyis metric independent. For $k>n$, the form space and the resulting class are zero. This proves metric independence of curvature Dolbeault powers without assuming that endomorphism-valued forms commute.

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.

Let $D_1$ and $D_2$ each be a connection on a vector bundle. Their difference is tensorial, so $D_1=D_2+a$ with $a\in {\Omega }^1(\mathrm{End}E)$. Extend $D_2$ to the endomorphism bundle connection. Expanding $(D_2+a{)}^2$ with the supplied graded Leibniz rule gives the curvature difference formula

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.