[proofplan] The proof identifies the Chern connection on the [tensor product](/page/Tensor%20Product) with the tensor product connection induced from the two Chern connections. We then compute the square of this connection on a local decomposable section $s\otimes t$. The only point needing care is the graded Leibniz rule: the mixed first-order terms have opposite signs and cancel, leaving exactly the two curvature terms. [/proofplan]

[step:Identify the Chern connection on the tensor product]For a smooth complex vector bundle $G\to X$, let $\Gamma(G)$ denote the space of smooth sections of $G$, and let $\Omega^1(X,G)$ denote the space of smooth $G$-valued $1$-forms on $X$. Let $\bar\partial_E$ and $\bar\partial_F$ denote the holomorphic structures on $E$ and $F$, respectively. Let $\nabla^E$ denote the Chern connection of $(E,h_E)$, and let $\nabla^F$ denote the Chern connection of $(F,h_F)$. Define a connection \begin{align*} \nabla^{E\otimes F}: \Gamma(E\otimes F) \to \Omega^1(X,E\otimes F) \end{align*} on decomposable smooth sections by \begin{align*} \nabla^{E\otimes F}(s\otimes t)=(\nabla^E s)\otimes t+s\otimes(\nabla^F t), \end{align*} where $s\in\Gamma(E)$ and $t\in\Gamma(F)$, and extend by linearity. This connection satisfies the Leibniz rule because both $\nabla^E$ and $\nabla^F$ do. Its $(0,1)$-part is \begin{align*} (\nabla^{E\otimes F})^{0,1}(s\otimes t)=(\bar\partial_E s)\otimes t+s\otimes(\bar\partial_F t), \end{align*} which is the holomorphic structure on $E\otimes F$. It is also compatible with the tensor [product metric](/page/Product%20Metric), because for local smooth sections $s_1,s_2\in\Gamma(E)$ and $t_1,t_2\in\Gamma(F)$, \begin{align*} d\bigl((h_E\otimes h_F)(s_1\otimes t_1,s_2\otimes t_2)\bigr) \end{align*} equals \begin{align*} d\bigl(h_E(s_1,s_2)h_F(t_1,t_2)\bigr), \end{align*} and the product rule together with metric compatibility of $\nabla^E$ and $\nabla^F$ gives \begin{align*} d\bigl((h_E\otimes h_F)(s_1\otimes t_1,s_2\otimes t_2)\bigr)=(h_E\otimes h_F)(\nabla^{E\otimes F}(s_1\otimes t_1),s_2\otimes t_2)+(h_E\otimes h_F)(s_1\otimes t_1,\nabla^{E\otimes F}(s_2\otimes t_2)). \end{align*} Indeed, the first term on the right contains $(h_E(\nabla^E s_1,s_2))h_F(t_1,t_2)+h_E(s_1,s_2)(h_F(\nabla^F t_1,t_2))$, and the second contains $(h_E(s_1,\nabla^E s_2))h_F(t_1,t_2)+h_E(s_1,s_2)(h_F(t_1,\nabla^F t_2))$, exactly matching the product-rule expansion. Hence, by the defining uniqueness of the Chern connection, this is the Chern connection of $(E\otimes F,h_E\otimes h_F)$.[/step]

[guided]We first need to justify that the connection whose curvature we compute is really the Chern connection on the tensor product. The Chern connection on a Hermitian holomorphic vector bundle is characterized by two properties: its $(0,1)$-part is the holomorphic $\bar\partial$-operator, and it is compatible with the Hermitian metric. For a smooth complex vector bundle $G\to X$, let $\Gamma(G)$ denote the space of smooth sections of $G$, and let $\Omega^1(X,G)$ denote the space of smooth $G$-valued $1$-forms. Let $\bar\partial_E$ and $\bar\partial_F$ denote the holomorphic structures on $E$ and $F$, respectively. Let $\nabla^E$ be the Chern connection of $(E,h_E)$ and $\nabla^F$ the Chern connection of $(F,h_F)$. Define \begin{align*} \nabla^{E\otimes F}: \Gamma(E\otimes F) \to \Omega^1(X,E\otimes F) \end{align*} on decomposable sections by \begin{align*} \nabla^{E\otimes F}(s\otimes t)=(\nabla^E s)\otimes t+s\otimes(\nabla^F t), \end{align*} where $s\in\Gamma(E)$ and $t\in\Gamma(F)$. Since local sections of $E\otimes F$ are locally finite sums of decomposable sections, this formula determines a connection after extending by linearity. Its $(0,1)$-part is \begin{align*} (\nabla^{E\otimes F})^{0,1}(s\otimes t)=(\bar\partial_E s)\otimes t+s\otimes(\bar\partial_F t). \end{align*} This is exactly the tensor product holomorphic structure on $E\otimes F$. Now check metric compatibility. For local smooth sections $s_1,s_2\in\Gamma(E)$ and $t_1,t_2\in\Gamma(F)$, the tensor product metric is defined by \begin{align*} (h_E\otimes h_F)(s_1\otimes t_1,s_2\otimes t_2)=h_E(s_1,s_2)h_F(t_1,t_2). \end{align*} Taking the [exterior derivative](/theorems/1525) and using the ordinary product rule gives \begin{align*} d\bigl(h_E(s_1,s_2)h_F(t_1,t_2)\bigr)=h_F(t_1,t_2)d h_E(s_1,s_2)+h_E(s_1,s_2)d h_F(t_1,t_2). \end{align*} Because $\nabla^E$ and $\nabla^F$ are metric-compatible, the two differentials on the right are exactly the pairings obtained by differentiating the $E$-factor and the $F$-factor. Substituting the definition of $\nabla^{E\otimes F}$ gives \begin{align*} d\bigl((h_E\otimes h_F)(s_1\otimes t_1,s_2\otimes t_2)\bigr)=(h_E\otimes h_F)(\nabla^{E\otimes F}(s_1\otimes t_1),s_2\otimes t_2)+(h_E\otimes h_F)(s_1\otimes t_1,\nabla^{E\otimes F}(s_2\otimes t_2)). \end{align*} The first pairing contributes $(h_E(\nabla^E s_1,s_2))h_F(t_1,t_2)+h_E(s_1,s_2)(h_F(\nabla^F t_1,t_2))$, while the second contributes $(h_E(s_1,\nabla^E s_2))h_F(t_1,t_2)+h_E(s_1,s_2)(h_F(t_1,\nabla^F t_2))$. These are precisely the four product-rule terms. Therefore $\nabla^{E\otimes F}$ has the two characterizing properties of the Chern connection, so by uniqueness it is the Chern connection on $(E\otimes F,h_E\otimes h_F)$.[/guided]

[step:Expand the square of the tensor product connection]Let $s\in\Gamma(E)$ and $t\in\Gamma(F)$ be local smooth sections. The connection $\nabla^{E\otimes F}$ extends to $E\otimes F$-valued forms by the graded Leibniz rule. When a $p$-form with values in $E$ is tensored with a $q$-form with values in $F$, the symbol $\otimes$ denotes the exterior wedge product of the form parts together with the tensor product of the bundle parts, giving an $(E\otimes F)$-valued $(p+q)$-form. Applying $\nabla^{E\otimes F}$ twice to $s\otimes t$ gives \begin{align*} (\nabla^{E\otimes F})^2(s\otimes t)=\nabla^{E\otimes F}\bigl((\nabla^E s)\otimes t+s\otimes(\nabla^F t)\bigr). \end{align*} For the first summand, $\nabla^E s$ has form degree $1$, so the graded sign is negative: \begin{align*} \nabla^{E\otimes F}\bigl((\nabla^E s)\otimes t\bigr)=(\nabla^E)^2s\otimes t-(\nabla^E s)\otimes(\nabla^F t). \end{align*} For the second summand, $s$ has form degree $0$, so no sign appears: \begin{align*} \nabla^{E\otimes F}\bigl(s\otimes(\nabla^F t)\bigr)=(\nabla^E s)\otimes(\nabla^F t)+s\otimes(\nabla^F)^2t. \end{align*} Adding the two identities cancels the mixed first-order terms, hence \begin{align*} (\nabla^{E\otimes F})^2(s\otimes t)=(\nabla^E)^2s\otimes t+s\otimes(\nabla^F)^2t. \end{align*}[/step]

[guided]Let $s\in\Gamma(E)$ and $t\in\Gamma(F)$ be local smooth sections. The [extension of a connection](/theorems/1539) to bundle-valued forms uses the graded Leibniz rule: when a form has odd degree, moving the connection past that form contributes a minus sign. Here $\nabla^E s$ is an $E$-valued $1$-form and $\nabla^F t$ is an $F$-valued $1$-form. In expressions such as $(\nabla^E s)\otimes(\nabla^F t)$, the form parts are wedged and the bundle parts are tensored. Applying the tensor product connection once gives \begin{align*} \nabla^{E\otimes F}(s\otimes t)=(\nabla^E s)\otimes t+s\otimes(\nabla^F t). \end{align*} Apply the same connection to the first summand. Since $\nabla^E s$ has form degree $1$, the graded Leibniz rule gives the negative sign in the mixed term: \begin{align*} \nabla^{E\otimes F}\bigl((\nabla^E s)\otimes t\bigr)=(\nabla^E)^2s\otimes t-(\nabla^E s)\otimes(\nabla^F t). \end{align*} Apply the connection to the second summand. Since $s$ has form degree $0$, no sign appears: \begin{align*} \nabla^{E\otimes F}\bigl(s\otimes(\nabla^F t)\bigr)=(\nabla^E s)\otimes(\nabla^F t)+s\otimes(\nabla^F)^2t. \end{align*} The two mixed terms are the same $(E\otimes F)$-valued $2$-form with opposite signs, so they cancel. Therefore \begin{align*} (\nabla^{E\otimes F})^2(s\otimes t)=(\nabla^E)^2s\otimes t+s\otimes(\nabla^F)^2t. \end{align*}[/guided]

[step:Identify the remaining terms as the two curvature summands] By definition of curvature, \begin{align*} \Theta(E,h_E)=(\nabla^E)^2,\qquad \Theta(F,h_F)=(\nabla^F)^2,\qquad \Theta(E\otimes F,h_E\otimes h_F)=(\nabla^{E\otimes F})^2. \end{align*} Therefore, for every local decomposable section $s\otimes t$, \begin{align*} \Theta(E\otimes F,h_E\otimes h_F)(s\otimes t)=\Theta(E,h_E)s\otimes t+s\otimes\Theta(F,h_F)t. \end{align*} This is precisely the action of \begin{align*} \Theta(E,h_E)\otimes \operatorname{id}_F+\operatorname{id}_E\otimes\Theta(F,h_F) \end{align*} on $s\otimes t$. Since decomposable local sections generate $E\otimes F$ locally, the two $\operatorname{End}(E\otimes F)$-valued $2$-forms are equal. Thus \begin{align*} \Theta(E\otimes F,h_E\otimes h_F)=\Theta(E,h_E)\otimes \operatorname{id}_F+\operatorname{id}_E\otimes\Theta(F,h_F). \end{align*} [/step]