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The curvature of a tensor product connection isFor line bundles, this is simply $F_{E\otimes F}=F_E+F_F$.

With a Hermitian form linear in its first slot, $H(s\otimes \bar{t})=h(s,t)$ is a complex-linear dual tensor. The original connection, its conjugate connection, the tensor product connection and the dual connection induce $D_0$. Its derivative is $(D_{0,V}H)(s\otimes \bar{t})=Vh(s,t)-h(D_{V}s,t)-h(s,D_{V}t)$. Consequently $D_{0}H=0$ is exactly metric compatibility.

Use the convention that a Hermitian metric $h$ is complex-linear in its first argument and conjugate-linear in its second. It is a smoothly varying positive-definite Hermitian form on each fibre. A connection on a vector bundle is a complex-linear operator $D:\Gamma (E)\to {\Omega }^1(X,E)$ satisfying $D(as)=da\otimes s+aDs$ for smooth complex functions $a$. Metric compatibility means, for every real vector field $V$,This is the metric-compatible connection condition with the sesquilinear convention fixed.

Apply the smooth Gram-Schmidt process to a local frame to obtain an $h$-orthonormal smooth frame $s_1,\dots ,s_r$. Positivity guarantees that all normalization denominators are nonzero and depend smoothly on the base point. Write $D{s}_j={\sum }_i{A}_{ij}{s}_i$. Differentiating $h(s_i,s_j)={\delta }_{ij}$ and using compatibility givesfor every real $V$. Thus $\boxed{A(V{)}^{†}=-A(V)}$: the connection matrix is skew-Hermitian. This smooth unitary frame for a Hermitian connection is generally not holomorphic; the requested local-frame assertion requires only a smooth frame.

The holomorphic dual vector bundle $E^{\ast }$ is obtained by dualizing fibres and using transition matrices $g_{ij}^{-T}$ when $E$ has transitions $g_{ij}$. Their entries are holomorphic because matrix inversion is holomorphic on $G{L}_r(\mathbb{C})$. Their cocycle property follows from preservation of the fibrewise evaluation pairing, defining the natural holomorphic bundle with fibre $(E_x{)}^{\ast }$.

The conjugate vector bundle $\overline{E}$ has the same underlying real fibres but opposite scalar action: $\lambda \cdot \overline{v}=\overline{\overline{\lambda }v}$. Its smooth transition matrices are $\overline{g_{ij}}$; they need not be holomorphic on $X$. The conjugate bundle is used here as a smooth complex bundle, whereas $E^{\ast }$ is holomorphic.

Define the smooth dual tensor $H$ byIt is complex-linear in each tensor factor, because conjugating the second bundle converts the conjugate-linearity of $h$ into linearity. Thus $H\in \Gamma ((E\otimes \overline{E}{)}^{\ast })$.

The conjugate connection is defined on real vector fields by ${\overline{D}}_V(\overline{s})=\overline{D_{V}s}$ and extended complex-linearly on the conjugate bundle. The tensor product connection isIts dual connection $D_0$ is uniquely characterized byThe Leibniz rule makes this a genuine connection on $(E\otimes \overline{E}{)}^{\ast }$. Evaluate it on the decomposable tensor $u=s\otimes \overline{t}$:Decomposable tensors span every fibre. Hence this tensor-valued one-form vanishes precisely when the compatibility identity holds for every $V,s,t$:This is metric compatibility as parallelism of a Hermitian tensor.

A connection on a vector bundle is a real-linear map (complex-linear for a complex bundle)It extends to the covariant exterior derivative by $D(\eta \otimes s)=d\eta \otimes s+(-1{)}^k\eta \wedge Ds$ when $\eta$ has degree $k$. Its square is linear over smooth functions: in $D^2(fs)$ the two $df\wedge Ds$ terms have opposite signs and $d^{2}f=0$. The curvature form of a connection is therefore the endomorphism-valued two-formIn a local frame with $D=d+\theta$, it is $\Theta =d\theta +\theta \wedge \theta$.

Define the tensor product connection on decomposable sections byThe Leibniz rule for $D_1,D_2$ makes this well-defined over smooth functions and makes it a connection on a vector bundle. Applying its covariant exterior derivative again givesThe mixed terms cancel by the graded sign. Thus curvature adds on the two tensor factors:Equivalently, the mixed terms in $({\theta }_1\otimes I+I\otimes {\theta }_2)\wedge ({\theta }_1\otimes I+I\otimes {\theta }_2)$ cancel because operators on the different factors commute. This also proves the curvature of a tensor product connection formula in local frames.

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.

A holomorphic line bundle is a complex line bundle with holomorphic transition functions, and a holomorphic section is one whose coefficient in every holomorphic local frame is holomorphic. Given a Hermitian metric on a holomorphic vector bundle $h$, its Chern connection is the connection $\nabla$ satisfying

Let $e$ be a nonvanishing holomorphic local frame, put $H=h(e,e)>0$, and write $\nabla e=Ae$. The first condition forces $A^{0,1}=0$, while metric compatibility forcesThis determines $\nabla$ uniquely and also constructs it. If $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $A^{\prime }=A+g^{-1}\partial g$, exactly the connection one-form transformation law, so the local constructions glue.

For a line bundle, $A\wedge A=0$, and the curvature form of a connection isIt has type $(1,1)$. Under $e^{\prime }=ge$, the extra term $g^{-1}\partial g$ is closed, so the curvature is unchanged and therefore global. This is the local formula for the Chern connection on a line bundle.

Any other Hermitian metric has the form $\hat{h}=e^{u}h$ for a global smooth real function $u$. Its local squared norm is $\hat{H}=e^{u}H$, whence

Connections ${\nabla }_1$ and ${\nabla }_2$ induce the tensor product connectionIts connection form in a product frame is $A_1+A_2$, so the curvature of a tensor product connection is $F_1+F_2$. For Chern connections, equip $L_1\otimes L_2$ with the product metricThe tensor product connection has the correct $(0,1)$ part and preserves this metric, so uniqueness identifies it with the Chern connection of $h$.

A real (1, 1)-form is a form $\phi \in {\Omega }^{1,1}(X)$ satisfying $\bar{\phi }=\phi$. In holomorphic coordinates it has the formIt is a positive real (1, 1)-form whenfor every nonzero tangent vector $\xi$ of type $(1,0)$, equivalently when the Hermitian matrix $(h_{j\bar{k}})$ is positive definite.

A holomorphic local trivialization of a holomorphic line bundle $L\to X$ is equivalently a nowhere-zero holomorphic local frame $e$. A connection $A$ is unitary when it preserves the fiberwise Hermitian inner product:The Chern connection is the unique unitary connection whose $(0,1)$ part is the bundle's Dolbeault partial connection ${\bar{\partial }}_L$.

In a holomorphic frame, put $h=h(e,e)$. The local formula for the Chern connection on a line bundle isThe curvature therefore has type $(1,1)$. Since a unitary connection has imaginary curvature, $\overline{F(A)}=-F(A)$, and henceThus $iF(A)$ is a real $(1,1)$-form.

For connections $A$ on $L$ and $\hat{A}$ on $\hat{L}$, the tensor product connection is defined on decomposable local sections byThe curvature of a tensor product connection on line bundles is additive:Consequentlywhich is positive whenever both summands are positive.

For the final assertion, simultaneously diagonalize the positive Hermitian matrices of $\phi$ and $\psi$ by congruence at the chosen point. In the resulting coframe,A direct wedge-product calculation givesIf $\xi$ and $\eta$ are linearly independent, at least one of these $2\times 2$ minors is nonzero. Every coefficient is positive, so the sum is strictly positive. This is wedge positivity for two positive (1, 1)-forms.