[proofplan]
We work locally in a holomorphic frame of $E$ and the corresponding dual frame of $E^*$. The Chern connection on the dual bundle is determined by differentiating the natural pairing, and this forces its connection matrix to be the negative fibre-index transpose of the connection matrix of $E$. Computing the two curvature matrices from the convention $\Theta=(\nabla)^2$ then gives the desired sign, with the extra sign in the wedge product coming from the fact that the connection matrix entries are $1$-forms.
[/proofplan]
[step:Choose a local holomorphic frame and define the connection matrices]
Let $U\subset X$ be an [open set](/page/Open%20Set) over which $E$ is holomorphically trivial, and let
\begin{align*}
(e_1,\dots,e_r)
\end{align*}
be a holomorphic frame for $E|_U$. Let
\begin{align*}
(e^1,\dots,e^r)
\end{align*}
be the dual frame for $E^*|_U$, so that $e^i(e_j)=\delta_{ij}$ for all $1\le i,j\le r$.
Define the matrix $A=(A_{ij})_{1\le i,j\le r}$ of complex-valued $1$-forms on $U$ by
\begin{align*}
\nabla^E e_j=\sum_{i=1}^r A_{ij}\otimes e_i.
\end{align*}
Define the matrix $B=(B_{ij})_{1\le i,j\le r}$ of complex-valued $1$-forms on $U$ by
\begin{align*}
\nabla^{E^*} e^j=\sum_{i=1}^r B_{ij}\otimes e^i.
\end{align*}
With the curvature convention $\Theta=(\nabla)^2$, the local curvature matrices are
\begin{align*}
\Theta(E,h)=dA+A\wedge A
\end{align*}
and
\begin{align*}
\Theta(E^*,h^*)=dB+B\wedge B,
\end{align*}
where
\begin{align*}
(A\wedge A)_{ij}=\sum_{k=1}^r A_{ik}\wedge A_{kj}
\end{align*}
and similarly for $B\wedge B$.
[/step]
[step:Use compatibility with the dual pairing to identify the dual connection matrix]
Let
\begin{align*}
s:U\to E|_U
\end{align*}
be a smooth local section and let
\begin{align*}
\lambda:U\to E^*|_U
\end{align*}
be a smooth local section of the dual bundle. The dual connection is characterized by the Leibniz rule for the natural pairing:
\begin{align*}
d(\lambda(s))=(\nabla^{E^*}\lambda)(s)+\lambda(\nabla^E s).
\end{align*}
Apply this identity to $\lambda=e^i$ and $s=e_j$. Since $e^i(e_j)=\delta_{ij}$ is constant, the left-hand side is zero. Therefore
\begin{align*}
0=(\nabla^{E^*}e^i)(e_j)+e^i(\nabla^E e_j).
\end{align*}
By the definitions of $A$ and $B$,
\begin{align*}
(\nabla^{E^*}e^i)(e_j)=B_{ji}
\end{align*}
and
\begin{align*}
e^i(\nabla^E e_j)=A_{ij}.
\end{align*}
Thus $B_{ji}+A_{ij}=0$ for all $i,j$, so
\begin{align*}
B=-A^t.
\end{align*}
[guided]
The purpose of this step is to determine the connection on $E^*$ from the connection on $E$. The defining condition is not guessed: it is forced by requiring the natural pairing between a covector and a vector to obey the ordinary [exterior derivative](/theorems/1525) rule.
Let
\begin{align*}
s:U\to E|_U
\end{align*}
be a smooth section and let
\begin{align*}
\lambda:U\to E^*|_U
\end{align*}
be a smooth dual section. The scalar function obtained by pairing them is
\begin{align*}
\lambda(s):U\to\mathbb C.
\end{align*}
The dual connection is defined so that
\begin{align*}
d(\lambda(s))=(\nabla^{E^*}\lambda)(s)+\lambda(\nabla^E s).
\end{align*}
Now apply this formula to the basis covector $e^i$ and the basis vector $e_j$. Their pairing is the constant function $\delta_{ij}$ on $U$, so
\begin{align*}
d(e^i(e_j))=d(\delta_{ij})=0.
\end{align*}
Hence
\begin{align*}
0=(\nabla^{E^*}e^i)(e_j)+e^i(\nabla^E e_j).
\end{align*}
By definition of the connection matrix $B$,
\begin{align*}
\nabla^{E^*}e^i=\sum_{k=1}^r B_{ki}\otimes e^k.
\end{align*}
Evaluating on $e_j$ gives
\begin{align*}
(\nabla^{E^*}e^i)(e_j)=\sum_{k=1}^r B_{ki}e^k(e_j)=B_{ji}.
\end{align*}
Similarly, by definition of the connection matrix $A$,
\begin{align*}
\nabla^E e_j=\sum_{k=1}^r A_{kj}\otimes e_k.
\end{align*}
Applying $e^i$ gives
\begin{align*}
e^i(\nabla^E e_j)=\sum_{k=1}^r A_{kj}e^i(e_k)=A_{ij}.
\end{align*}
Substituting these two identities into the pairing rule yields
\begin{align*}
B_{ji}+A_{ij}=0.
\end{align*}
Since this holds for every pair of indices $i,j$, the matrix of the dual connection is
\begin{align*}
B=-A^t.
\end{align*}
[/guided]
[/step]
[step:Compute the curvature matrix of the dual connection]
Using $B=-A^t$, we compute
\begin{align*}
dB=-d(A^t)=-(dA)^t.
\end{align*}
For the quadratic term, the entries of $B\wedge B$ are
\begin{align*}
(B\wedge B)_{ij}=\sum_{k=1}^r B_{ik}\wedge B_{kj}.
\end{align*}
Since $B_{ik}=-A_{ki}$ and $B_{kj}=-A_{jk}$, this becomes
\begin{align*}
(B\wedge B)_{ij}=\sum_{k=1}^r A_{ki}\wedge A_{jk}.
\end{align*}
Each $A_{ki}$ and $A_{jk}$ is a $1$-form, so graded commutativity gives
\begin{align*}
A_{ki}\wedge A_{jk}=-A_{jk}\wedge A_{ki}.
\end{align*}
Therefore
\begin{align*}
(B\wedge B)_{ij}=-\sum_{k=1}^r A_{jk}\wedge A_{ki}.
\end{align*}
The right-hand side is exactly $-(A\wedge A)^t_{ij}$. Hence
\begin{align*}
B\wedge B=-(A\wedge A)^t.
\end{align*}
Combining the two terms,
\begin{align*}
\Theta(E^*,h^*)=dB+B\wedge B=-(dA)^t-(A\wedge A)^t=-(dA+A\wedge A)^t.
\end{align*}
Thus, in the chosen frame,
\begin{align*}
\Theta(E^*,h^*)=-\Theta(E,h)^t.
\end{align*}
[/step]
[step:Translate the local matrix identity into the intrinsic pairing formula]
The transpose in the displayed local formula is the transpose on fibre indices. Therefore, for every $x\in U$, every $\xi,\eta\in T_xX$, every $\lambda\in E_x^*$, and every $s\in E_x$, the local matrix identity gives
\begin{align*}
\bigl(\Theta(E^*,h^*)(\xi,\eta)\lambda\bigr)(s)=-\lambda\bigl(\Theta(E,h)(\xi,\eta)s\bigr).
\end{align*}
This identity is independent of the chosen holomorphic frame because it is expressed entirely through the natural pairing $E_x^*\times E_x\to\mathbb C$. Since the open set $U$ was arbitrary, the identity holds on all of $X$. Equivalently,
\begin{align*}
\Theta(E^*,h^*)=-{}^t\Theta(E,h).
\end{align*}
This proves the curvature formula for the dual bundle.
[/step]
