[proofplan] We use the Hodge metric to split the Hodge filtration smoothly and identify each graded bundle $E^p$ with its orthogonal representative inside $\mathcal V$. With respect to this splitting, the flat Gauss--Manin connection decomposes into the direct sum of the Chern connections on the graded pieces, the Higgs field $\theta$, and its metric adjoint $\theta^*$. Taking the $E^p$ diagonal block of the flatness identity $(\nabla)^2=0$ gives exactly the curvature formula, because the only two-step paths returning to $E^p$ are the outgoing term through $E^{p-1}$ and the incoming term through $E^{p+1}$. For $p=k$, the incoming term vanishes, so the curvature is a negative square, which is precisely Griffiths seminegativity. [/proofplan]

[step:Split the Hodge filtration by the Hodge metric] Let $h$ denote the Hermitian Hodge metric on the $C^\infty$ complex vector bundle underlying $\mathcal V$. For each $p$, let $H^p\subset F^p$ denote the $h$-orthogonal complement of $F^{p+1}$ in $F^p$. The quotient map $F^p\to F^p/F^{p+1}=E^p$ restricts to a smooth Hermitian vector bundle isomorphism \begin{align*} H^p\longrightarrow E^p. \end{align*} Using this isomorphism, we regard $E^p$ as the smooth subbundle $H^p\subset \mathcal V$. Thus \begin{align*} \mathcal V=\bigoplus_{r=0}^k E^r \end{align*} as a smooth Hermitian vector bundle. Let \begin{align*} D:=\bigoplus_{r=0}^k \nabla_{E^r} \end{align*} be the direct sum of the Chern connections on the Hermitian holomorphic bundles $E^r$. Here $\nabla_{E^r}$ is the unique connection on $E^r$ whose $(0,1)$-part is the holomorphic structure and which is compatible with the Hodge metric. [/step]

[step:Identify the off-diagonal connection terms with the Higgs field and its adjoint]With respect to the smooth [orthogonal decomposition](/theorems/436) $\mathcal V=\bigoplus_{r=0}^k E^r$, [Griffiths transversality](/theorems/9129) says that the $(1,0)$-part of $\nabla$ maps $F^p$ into $F^{p-1}\otimes\Omega_S^{1,0}$. Therefore its induced off-diagonal component from $E^p$ to $E^{p-1}$ is exactly \begin{align*} \theta_p\in \Omega_S^{1,0}(\operatorname{Hom}(E^p,E^{p-1})). \end{align*} The polarization makes $\nabla$ compatible with the Hodge metric in the standard Hodge-theoretic sense, so the opposite off-diagonal component is the metric adjoint \begin{align*} \theta_p^*\in \Omega_S^{0,1}(\operatorname{Hom}(E^{p-1},E^p)). \end{align*} Consequently, as a connection on the smooth bundle $\bigoplus_{r=0}^kE^r$, \begin{align*} \nabla=D+\theta+\theta^*, \end{align*} where \begin{align*} \theta:=\sum_{r=1}^k\theta_r \end{align*} lowers the Hodge index by one and \begin{align*} \theta^*:=\sum_{r=1}^k\theta_r^* \end{align*} raises the Hodge index by one.[/step]

[guided]The point of introducing the Hodge metric splitting is that the filtered bundle $\mathcal V$ becomes a direct sum of the graded pieces $E^r$ at the $C^\infty$ level. This lets us write the connection as a matrix whose rows and columns are indexed by the Hodge degree. The diagonal part of this matrix is the direct sum \begin{align*} D:=\bigoplus_{r=0}^k\nabla_{E^r}, \end{align*} where $\nabla_{E^r}$ is the Chern connection of the Hermitian holomorphic bundle $E^r$. The off-diagonal part is controlled by Griffiths transversality. Namely, the $(1,0)$-part of the Gauss--Manin connection sends $F^p$ into $F^{p-1}\otimes\Omega_S^{1,0}$, so after passing to the quotient $F^p/F^{p+1}$ it has only one possible lowering component: \begin{align*} \theta_p:E^p\longrightarrow E^{p-1}\otimes\Omega_S^{1,0}. \end{align*} This is precisely the Higgs component in the statement. Because the splitting is orthogonal for the Hodge metric, the corresponding raising component is not an independent object. It is the Hermitian adjoint \begin{align*} \theta_p^*:E^{p-1}\longrightarrow E^p\otimes\Omega_S^{0,1}. \end{align*} Thus the full connection has the block form \begin{align*} \nabla=D+\theta+\theta^*, \end{align*} where $\theta=\sum_{r=1}^k\theta_r$ lowers Hodge degree by one and $\theta^*=\sum_{r=1}^k\theta_r^*$ raises Hodge degree by one. This decomposition is the whole mechanism behind the curvature formula: curvature is obtained by squaring this connection and looking at the diagonal block.[/guided]

[step:Take the degree-preserving block of the flatness equation] The Gauss--Manin connection is flat, so \begin{align*} 0=(\nabla)^2. \end{align*} Using \begin{align*} \nabla=D+\theta+\theta^*, \end{align*} we take the component of $(\nabla)^2$ that maps $E^p$ to $E^p$ and has form type $(1,1)$. The diagonal contribution is \begin{align*} D^2|_{E^p}=\Theta_{E^p}. \end{align*} Since $\theta$ lowers the Hodge index by one and $\theta^*$ raises it by one, the only off-diagonal two-step contributions from $E^p$ back to $E^p$ are \begin{align*} \theta_p^*\wedge\theta_p \end{align*} and \begin{align*} -\theta_{p+1}\wedge\theta_{p+1}^*. \end{align*} The minus sign is the exterior sign from commuting a $(0,1)$-form past a $(1,0)$-form under the stated wedge-composition convention. Hence the $E^p$ diagonal block of $(\nabla)^2=0$ is \begin{align*} 0=\Theta_{E^p}+\theta_p^*\wedge\theta_p-\theta_{p+1}\wedge\theta_{p+1}^*. \end{align*} Rearranging gives \begin{align*} \Theta_{E^p}=\theta_{p+1}\wedge\theta_{p+1}^*-\theta_p^*\wedge\theta_p. \end{align*} This proves the curvature identity for every $0\le p\le k$, with the convention $\theta_p=0$ outside $\{1,\dots,k\}$. [/step]

[step:Read the top Hodge bundle curvature as a negative square] For $p=k$, the convention $\theta_{k+1}=0$ gives \begin{align*} \Theta_{E^k}=-\theta_k^*\wedge\theta_k. \end{align*} Since $F^{k+1}=0$, one has $F^k=F^k/F^{k+1}=E^k$. Let $s\in (F^k)_x$ and let $\xi\in T_x^{1,0}S$. Evaluating the Hermitian curvature form on $\xi$ and $\bar{\xi}$ gives \begin{align*} h\bigl(i\Theta_{F^k}(\xi,\bar{\xi})s,s\bigr)=-\,|\theta_k(\xi)s|_h^2\le 0. \end{align*} Thus $i\Theta_{F^k}$ is Griffiths seminegative. The curvature of the dual Hermitian bundle is the negative transpose adjoint of the original curvature, so dualizing reverses Griffiths seminegativity into Griffiths semipositivity. Therefore $(F^k)^*$ is Griffiths semipositive. [/step]