[proofplan] The Chern connection on $E$ induces a connection on $E^{\otimes m}$ by differentiating each tensor factor, and the symmetric subbundle is preserved because the construction is invariant under permutations of the tensor factors. Squaring this induced connection gives the infinitesimal action of the curvature endomorphism on symmetric tensors. To prove Griffiths semipositivity, fix a tangent direction, diagonalize the corresponding Hermitian curvature endomorphism on $E_x$, and observe that the induced operator on $S^mE_x$ is diagonal with eigenvalues given by sums of nonnegative eigenvalues. [/proofplan]

[step:Construct the induced Chern connection on symmetric tensors] Let $\nabla^E$ denote the Chern connection of $(E,h)$. Let $T^*X$ denote the cotangent bundle of $X$. For a smooth vector bundle $F\to X$, let $\Gamma(F)$ denote the [vector space](/page/Vector%20Space) of smooth sections of $F$. Define a connection $\nabla^{\otimes m}:\Gamma(E^{\otimes m})\to \Gamma(T^*X\otimes E^{\otimes m})$ on decomposable smooth sections by \begin{align*} \nabla^{\otimes m}(s_1\otimes \cdots \otimes s_m) = \sum_{a=1}^{m} s_1\otimes \cdots \otimes \nabla^E s_a\otimes \cdots \otimes s_m, \end{align*} where $s_1,\dots,s_m\in \Gamma(E)$ are smooth sections, and extend by linearity. Let $S_m$ denote the [symmetric group](/page/Symmetric%20Group) on the set $\{1,\dots,m\}$, acting on $E^{\otimes m}$ by permuting tensor factors. The formula is invariant under this action of $S_m$, so $\nabla^{\otimes m}$ preserves the symmetric subbundle $S^mE\subset E^{\otimes m}$. Let $\nabla^{S^mE}:\Gamma(S^mE)\to \Gamma(T^*X\otimes S^mE)$ denote the restricted connection. Because $\nabla^E$ is compatible with $h$, the [tensor product](/page/Tensor%20Product) connection is compatible with the tensor [product metric](/page/Product%20Metric) on $E^{\otimes m}$. Its restriction to $S^mE$ is therefore compatible with $h_{S^mE}$. Since $\nabla^E$ has $(0,1)$-part equal to the holomorphic structure operator $\bar\partial_E$, the induced connection has $(0,1)$-part equal to the induced holomorphic structure operator on $S^mE$. Hence $\nabla^{S^mE}$ is the Chern connection of $(S^mE,h_{S^mE})$. [/step]

[step:Compute the curvature as the infinitesimal symmetric power action]Let $\Omega^2(X,\operatorname{End}(E))$ denote the vector space of smooth two-forms on $X$ with values in $\operatorname{End}(E)$. Let $R^E=(\nabla^E)^2\in \Omega^2(X,\operatorname{End}(E))$ be the curvature two-form of $\nabla^E$. The bundles $(E,h)$ and $(E,h)$ are Hermitian holomorphic vector bundles on $X$, so the hypotheses of the [tensor product curvature formula](/theorems/9111) [citetheorem:9111] apply to each tensor factor in the iterated tensor product $E^{\otimes m}$. Applying that formula repeatedly gives the curvature of the tensor product connection. On decomposable tensors, \begin{align*} R^{\otimes m}(s_1\otimes \cdots \otimes s_m) = \sum_{a=1}^{m} s_1\otimes \cdots \otimes R^E s_a\otimes \cdots \otimes s_m. \end{align*} The right-hand side is symmetric whenever $s_1\otimes\cdots\otimes s_m$ is symmetrized, so restricting to $S^mE$ gives \begin{align*} R^{S^mE}=\rho_m(R^E). \end{align*} For the Chern curvature notation $\Theta$, this is exactly \begin{align*} \Theta(S^mE,h_{S^mE})=\rho_m\bigl(\Theta(E,h)\bigr). \end{align*}[/step]

[guided]We now compute what happens when the induced connection is squared. The point of using the tensor product connection is that curvature behaves by the same Leibniz principle as the connection itself: a curvature endomorphism can hit any one tensor factor, and the other tensor factors remain unchanged. Let $\Omega^2(X,\operatorname{End}(E))$ denote the vector space of smooth two-forms on $X$ with values in $\operatorname{End}(E)$. Let $R^E=(\nabla^E)^2\in \Omega^2(X,\operatorname{End}(E))$ be the curvature two-form of the Chern connection on $E$. The tensor product curvature formula [citetheorem:9111] applies because each tensor factor is the Hermitian holomorphic vector bundle $(E,h)$ on the same complex manifold $X$, and the metric on $E^{\otimes m}$ is the induced tensor product metric. Applying that formula repeatedly to the bundle $E^{\otimes m}$ gives the curvature \begin{align*} R^{\otimes m}\in \Omega^2(X,\operatorname{End}(E^{\otimes m})). \end{align*} On a decomposable tensor $s_1\otimes\cdots\otimes s_m$, where each $s_a\in \Gamma(E)$ is a smooth local section, the formula is \begin{align*} R^{\otimes m}(s_1\otimes \cdots \otimes s_m) = \sum_{a=1}^{m} s_1\otimes \cdots \otimes R^E s_a\otimes \cdots \otimes s_m. \end{align*} This formula is exactly the infinitesimal representation of $\operatorname{End}(E)$ on the $m$-fold tensor power: an endomorphism acts by being inserted into one tensor factor at a time and then summed over all tensor positions. Because the expression is invariant under permuting the tensor factors, it maps symmetric tensors to symmetric tensors. Therefore, after restricting from $E^{\otimes m}$ to the symmetric subbundle $S^mE$, the curvature is \begin{align*} R^{S^mE}=\rho_m(R^E). \end{align*} Finally, the symbol $\Theta$ denotes the same Chern curvature form in the holomorphic Hermitian bundle notation. Replacing $R$ by $\Theta$ gives \begin{align*} \Theta(S^mE,h_{S^mE})=\rho_m\bigl(\Theta(E,h)\bigr). \end{align*} This proves the asserted curvature formula.[/guided]

[step:Diagonalize the curvature endomorphism in one tangent direction] Fix $x\in X$ and $\xi\in T_xX$. Define the Hermitian endomorphism \begin{align*} A_{\xi}:E_x\to E_x \end{align*} by \begin{align*} A_{\xi}=i\Theta(E,h)(\xi,\overline{\xi}). \end{align*} If $(E,h)$ is Griffiths semipositive, then \begin{align*} (A_{\xi}v,v)_h\ge 0 \end{align*} for every $v\in E_x$. Since $A_{\xi}$ is Hermitian, choose an $h$-[orthonormal basis](/page/Orthonormal%20Basis) $e_1,\dots,e_r$ of $E_x$ consisting of eigenvectors of $A_{\xi}$. Let $\lambda_1,\dots,\lambda_r\in \mathbb R$ be the corresponding eigenvalues, so \begin{align*} A_{\xi}e_j=\lambda_j e_j \end{align*} for each $j\in\{1,\dots,r\}$. The Griffiths semipositivity inequality gives $\lambda_j\ge 0$ for every $j$ by testing it on $v=e_j$. [/step]

[step:Read the induced eigenvalues on the symmetric power] For a multi-index $\alpha=(\alpha_1,\dots,\alpha_r)\in \mathbb N_0^r$ with \begin{align*} \alpha_1+\cdots+\alpha_r=m, \end{align*} let \begin{align*} e^{\alpha}\in S^mE_x \end{align*} denote the normalized symmetrized tensor with $\alpha_j$ occurrences of $e_j$, scaled to have $h_{S^mE}$-norm $1$. These vectors form an orthonormal basis of $S^mE_x$ for the induced Hermitian metric. By the definition of $\rho_m$, the operator $\rho_m(A_{\xi})$ acts on this basis vector by applying $A_{\xi}$ to each tensor factor and summing the results. Therefore \begin{align*} \rho_m(A_{\xi})e^{\alpha} = \left(\sum_{j=1}^{r}\alpha_j\lambda_j\right)e^{\alpha}. \end{align*} Since each $\alpha_j\ge 0$ and each $\lambda_j\ge 0$, the eigenvalue \begin{align*} \sum_{j=1}^{r}\alpha_j\lambda_j \end{align*} is nonnegative. [/step]

[step:Conclude Griffiths semipositivity of the symmetric power] Let $u\in S^mE_x$. Expand $u$ in the orthonormal symmetric monomial basis as \begin{align*} u=\sum_{\alpha}u_{\alpha}e^{\alpha}, \end{align*} where the sum is over all $\alpha\in \mathbb N_0^r$ satisfying $\alpha_1+\cdots+\alpha_r=m$, and where $u_{\alpha}\in \mathbb C$. Using the diagonal action computed above, \begin{align*} \bigl(i\Theta(S^mE,h_{S^mE})(\xi,\overline{\xi})u,u\bigr)_{h_{S^mE}} = \sum_{\alpha} \left(\sum_{j=1}^{r}\alpha_j\lambda_j\right)|u_{\alpha}|^2. \end{align*} Every summand is nonnegative, so \begin{align*} \bigl(i\Theta(S^mE,h_{S^mE})(\xi,\overline{\xi})u,u\bigr)_{h_{S^mE}}\ge 0. \end{align*} Since $x\in X$, $\xi\in T_xX$, and $u\in S^mE_x$ were arbitrary, $(S^mE,h_{S^mE})$ is Griffiths semipositive. This completes the proof. [/step]