[proofplan]
We use the Hermitian metric to replace the quotient bundle by the orthogonal complement of its kernel. In this splitting, the Chern connection of $E$ decomposes into diagonal Chern connections on the subbundle and quotient, together with off-diagonal second fundamental form terms. Computing the curvature matrix and taking the $Q$-diagonal block gives the quotient curvature formula: ambient curvature restricted to the horizontal lift plus a squared norm term. Griffiths semipositivity and positivity then follow by evaluating this identity on tangent and fibre vectors.
[/proofplan]
[step:Identify the quotient bundle with the orthogonal complement of its kernel]
Let $S=\ker q\subset E$. Since $q:E\to Q$ is a holomorphic quotient bundle, $S$ is a holomorphic subbundle of $E$, and there is a short exact sequence of holomorphic vector bundles
\begin{align*}
0\longrightarrow S\longrightarrow E\xrightarrow{q}Q\longrightarrow 0.
\end{align*}
For each $x\in X$, let $S_x^\perp\subset E_x$ denote the $h_E$-orthogonal complement of $S_x$. The restriction
$q_x|_{S_x^\perp}:S_x^\perp\to Q_x$
is an isomorphism of complex vector spaces because its kernel is $S_x\cap S_x^\perp=\{0\}$ and its image has dimension $\operatorname{rank}Q$. The quotient metric $h_Q$ is exactly the metric for which this isomorphism is an isometry. Thus we may regard $Q$ as the smooth Hermitian subbundle $S^\perp\subset E$, while remembering that the holomorphic structure on $Q$ is the quotient holomorphic structure.
For $u\in Q_x$, denote by $\tilde u\in S_x^\perp$ the unique vector satisfying $q_x(\tilde u)=u$.
[/step]
[step:Decompose the Chern connection relative to the orthogonal splitting]
Let $D_E$ be the Chern connection of $(E,h_E)$. With respect to the smooth orthogonal splitting
\begin{align*}
E=S\oplus S^\perp,
\end{align*}
write a smooth section $e\in \Gamma(E)$ as $e=s+v$, where $\Gamma(E)$ denotes the space of smooth sections of $E\to X$, $s\in\Gamma(S)$, and $v\in\Gamma(S^\perp)$. Let $P_S:E\to S$ and $P_Q:E\to S^\perp$ be the smooth $h_E$-orthogonal projections.
Define the connection $D_S$ on $S$ by
\begin{align*}
D_S s=P_S(D_Es)
\end{align*}
for $s\in\Gamma(S)$. Define the connection $D_Q$ on the smooth bundle $S^\perp$ by
\begin{align*}
D_Q v=P_Q(D_Ev)
\end{align*}
for $v\in\Gamma(S^\perp)$. Under the identification $Q\cong S^\perp$, this $D_Q$ is the Chern connection of $(Q,h_Q)$. It is metric because $D_E$ is metric and $P_Q$ is the [orthogonal projection](/theorems/437). Its $(0,1)$-part is the quotient holomorphic structure: if $\sigma$ is a local smooth section of $Q$ and $\tilde\sigma$ is any local smooth lift to $E$, then $\bar\partial_Q\sigma$ is represented by $P_Q(\bar\partial_E\tilde\sigma)$; this is independent of the lift because the difference of two lifts is a section of the holomorphic subbundle $S$.
Define the second fundamental form of $S\subset E$ as the $(1,0)$-form with values in $\operatorname{Hom}(S,S^\perp)$ given by
\begin{align*}
B(s)=P_Q(D_Es)
\end{align*}
for $s\in\Gamma(S)$. Let $B^*$ denote its Hermitian adjoint $(0,1)$-form with values in $\operatorname{Hom}(S^\perp,S)$, so that for every $\xi\in T_x^{1,0}X$, every $s\in S_x$, and every $v\in S_x^\perp$,
\begin{align*}
h_E(B_\xi s,v)=h_E(s,B^*_{\bar\xi}v).
\end{align*}
Because $D_E$ is compatible with $h_E$ and the splitting is orthogonal, its action on decomposed sections is
\begin{align*}
D_E(s+v)=(D_Ss-B^*v)+(Bs+D_Qv),
\end{align*}
where the first parenthesis lies in $S$ and the second parenthesis lies in $S^\perp$. Here $B^*v$ denotes the $S$-valued one-form obtained by applying the $\operatorname{Hom}(S^\perp,S)$-valued $(0,1)$-form $B^*$ to the section $v$; explicitly, $(B^*v)_{\bar\eta}=B^*_{\bar\eta}v$ for $\eta\in T^{1,0}X$.
[guided]
The quotient metric is designed so that vectors in $Q_x$ are represented by their shortest lifts to $E_x$. Those shortest lifts are precisely the vectors orthogonal to $S_x=\ker q_x$. This is why the proof replaces the quotient bundle by the smooth bundle $S^\perp$.
Let $P_S:E\to S$ and $P_Q:E\to S^\perp$ be the two orthogonal projections. The Chern connection $D_E$ differentiates sections of $E$, but after differentiating a section of $S$ the result need not remain in $S$. Its $S^\perp$-part measures the failure of $S$ to be parallel. This is the second fundamental form
\begin{align*}
B:\Gamma(S)\to \Gamma(T^*X\otimes S^\perp),\qquad B(s)=P_Q(D_Es).
\end{align*}
Similarly, if $v$ is a section of $S^\perp$, then $P_Q(D_Ev)$ is the induced connection on the quotient side:
\begin{align*}
D_Qv=P_Q(D_Ev).
\end{align*}
This is the Chern connection of the quotient metric because it preserves the induced metric and its $(0,1)$-part agrees with the holomorphic quotient structure.
The off-diagonal term from $S^\perp$ back to $S$ is the negative metric adjoint of $B$. Indeed, if $s\in\Gamma(S)$ and $v\in\Gamma(S^\perp)$, then $h_E(s,v)=0$. Applying the metric connection $D_E$ to this identity in a tangent direction $\eta$ gives
\begin{align*}
0=\eta\,h_E(s,v)=h_E((D_E)_\eta s,v)+h_E(s,(D_E)_\eta v).
\end{align*}
The first term contains $h_E(B_\eta s,v)$, while the second term contains the $S$-component of $(D_E)_\eta v$. Therefore that $S$-component is $-B_\eta^*v$, where $B_\eta^*$ is defined by
\begin{align*}
h_E(B_\eta s,v)=h_E(s,B_\eta^*v).
\end{align*}
Thus, for every local decomposition $e=s+v$ with $s\in\Gamma(S)$ and $v\in\Gamma(S^\perp)$, the connection decomposes as
\begin{align*}
D_E(s+v)=(D_Ss-B^*v)+(Bs+D_Qv),
\end{align*}
with the first term in $S$ and the second term in $S^\perp$.
[/guided]
[/step]
[step:Compute the quotient curvature block]
Let $\Theta_E=D_E^2$, $\Theta_S=D_S^2$, and $\Theta_Q=D_Q^2$ denote the Chern curvatures of $E$, $S$, and $Q\cong S^\perp$. Let $(\Theta_E)_{Q,Q}$ denote the $S^\perp\to S^\perp$ diagonal block of $\Theta_E$ with respect to $E=S\oplus S^\perp$.
In block form the connection has diagonal entries $D_S,D_Q$ and off-diagonal entries $-B^*,B$. The $S^\perp\to S^\perp$ diagonal block of $D_E^2$ is obtained by composing the two paths that start and end in $S^\perp$: the diagonal path gives $D_Q^2=\Theta_Q$, and the off-diagonal path gives $B\wedge(-B^*)=-B\wedge B^*$ because connection forms are multiplied by exterior composition. Hence
\begin{align*}
(\Theta_E)_{Q,Q}=\Theta_Q-B\wedge B^*.
\end{align*}
Equivalently,
\begin{align*}
\Theta_Q=(\Theta_E)_{Q,Q}+B\wedge B^*.
\end{align*}
Fix $x\in X$, $\xi\in T_x^{1,0}X$, and $u\in Q_x$, and let $\tilde u\in S_x^\perp$ be the unique lift of $u$. Define the quotient-to-subspace off-diagonal second fundamental form in the direction $\xi$ by
\begin{align*}
A_\xi u:=B^*_{\bar\xi}\tilde u\in S_x.
\end{align*}
By the adjoint convention, $B^*_{\bar\xi}:S_x^\perp\to S_x$ is the Hermitian adjoint of $B_\xi:S_x\to S_x^\perp$. With the stated curvature convention, the adjoint second fundamental form term satisfies
\begin{align*}
h_E\bigl(i(B\wedge B^*)(\xi,\bar\xi)\tilde u,\tilde u\bigr)=h_E(B_\xi B^*_{\bar\xi}\tilde u,\tilde u)=h_E(B^*_{\bar\xi}\tilde u,B^*_{\bar\xi}\tilde u)=\|A_\xi u\|_{h_E}^2.
\end{align*}
Thus the sign is the positive sign in $\Theta_Q=(\Theta_E)_{Q,Q}+B\wedge B^*$. Therefore
\begin{align*}
h_Q\bigl(i\Theta(Q,h_Q)(\xi,\bar\xi)u,u\bigr)
=
h_E\bigl(i\Theta(E,h_E)(\xi,\bar\xi)\tilde u,\tilde u\bigr)
+
\|A_\xi u\|_{h_E}^2.
\end{align*}
The last term is nonnegative because it is the squared norm of a vector in the Hermitian space $S_x$.
[guided]
The only delicate point in the proof is the sign of the second fundamental form term. The form $B$ is the $(1,0)$ second fundamental form of the holomorphic subbundle $S\subset E$, so $B_\xi:S_x\to S_x^\perp$ for $\xi\in T_x^{1,0}X$. Its Hermitian adjoint is evaluated in the conjugate direction:
\begin{align*}
B^*_{\bar\xi}:S_x^\perp\to S_x.
\end{align*}
This means that for $s\in S_x$ and $v\in S_x^\perp$,
\begin{align*}
h_E(B_\xi s,v)=h_E(s,B^*_{\bar\xi}v).
\end{align*}
With respect to the splitting $E=S\oplus S^\perp$, the $S^\perp\to S^\perp$ diagonal block of the ambient curvature is denoted by $(\Theta_E)_{Q,Q}$. The connection decomposition says that a section of $S^\perp$ can return to $S^\perp$ in two ways after applying $D_E$ twice. The diagonal path is $D_Q^2=\Theta_Q$. The off-diagonal path first goes from $S^\perp$ to $S$ by $-B^*$ and then from $S$ to $S^\perp$ by $B$, so exterior composition gives the contribution $B\wedge(-B^*)=-B\wedge B^*$. Therefore
\begin{align*}
(\Theta_E)_{Q,Q}=\Theta_Q-B\wedge B^*.
\end{align*}
Solving for the quotient curvature gives
\begin{align*}
\Theta_Q=(\Theta_E)_{Q,Q}+B\wedge B^*.
\end{align*}
Now take $u\in Q_x$ and let $\tilde u\in S_x^\perp$ be its orthogonal lift. Define
\begin{align*}
A_\xi u:=B^*_{\bar\xi}\tilde u\in S_x.
\end{align*}
Under the stated convention for testing Chern curvature by $i\Theta(\xi,\bar\xi)$, the second fundamental form contribution is evaluated as
\begin{align*}
h_E\bigl(i(B\wedge B^*)(\xi,\bar\xi)\tilde u,\tilde u\bigr)=h_E(B_\xi B^*_{\bar\xi}\tilde u,\tilde u).
\end{align*}
The adjoint identity then converts this pairing into the norm square
\begin{align*}
h_E(B_\xi B^*_{\bar\xi}\tilde u,\tilde u)=h_E(B^*_{\bar\xi}\tilde u,B^*_{\bar\xi}\tilde u)=\|A_\xi u\|_{h_E}^2.
\end{align*}
Thus the quotient curvature satisfies
\begin{align*}
h_Q\bigl(i\Theta(Q,h_Q)(\xi,\bar\xi)u,u\bigr)
=
h_E\bigl(i\Theta(E,h_E)(\xi,\bar\xi)\tilde u,\tilde u\bigr)
+
\|A_\xi u\|_{h_E}^2.
\end{align*}
The second term is nonnegative because it is a Hermitian squared norm.
[/guided]
[/step]
[step:Conclude Griffiths semipositivity and positivity]
Assume $(E,h_E)$ is Griffiths semipositive. By the definition of Griffiths semipositivity, for every $x\in X$, every $\xi\in T_x^{1,0}X$, and every $w\in E_x$,
\begin{align*}
h_E\bigl(i\Theta(E,h_E)(\xi,\bar\xi)w,w\bigr)\ge 0.
\end{align*}
Applying this with $w=\tilde u$ and using the quotient curvature identity gives
\begin{align*}
h_Q\bigl(i\Theta(Q,h_Q)(\xi,\bar\xi)u,u\bigr)\ge 0.
\end{align*}
Thus $(Q,h_Q)$ is Griffiths semipositive.
If $(E,h_E)$ is Griffiths positive, $\xi\ne0$, and $u\ne0$, then $\tilde u\ne0$ because $q_x|_{S_x^\perp}$ is an isomorphism. Hence
\begin{align*}
h_E\bigl(i\Theta(E,h_E)(\xi,\bar\xi)\tilde u,\tilde u\bigr)>0.
\end{align*}
The second fundamental form contribution $\|A_\xi u\|_{h_E}^2$ is nonnegative, so
\begin{align*}
h_Q\bigl(i\Theta(Q,h_Q)(\xi,\bar\xi)u,u\bigr)>0.
\end{align*}
This proves both the descent of Griffiths semipositivity to quotient bundles and the stated positivity assertion in nonzero tested fibre directions.
[/step]
