[proofplan]
We prove the equivalence in the local unitary frame. In one direction, metric compatibility applied to the basis sections $e_i$ and $e_j$ differentiates the identity $h(e_i,e_j)=\delta_{ij}$ and gives the coefficient relation $A_{ji}+\overline{A_{ij}}=0$. This is exactly the entrywise form of $A^*+A=0$. Conversely, if $A$ is skew-Hermitian, we expand arbitrary sections in the unitary frame, compute both sides of the metric-compatibility identity, and use the skew-Hermitian coefficient relation to cancel the connection terms.
[/proofplan]
[step:Derive the skew-Hermitian coefficient identity from metric compatibility]
Assume first that $\nabla$ is compatible with $h$ on $U$. Fix indices $i,j \in \{1,\dots,r\}$. Since $e$ is a unitary frame, the function
\begin{align*}
h(e_i,e_j):U \to \mathbb{C}
\end{align*}
is the constant function $\delta_{ij}$, hence
\begin{align*}
d(h(e_i,e_j))=0.
\end{align*}
Metric compatibility applied to the sections $e_i,e_j \in \Gamma(U,E)$ gives
\begin{align*}
0=h(\nabla e_i,e_j)+h(e_i,\nabla e_j).
\end{align*}
By the definition of the connection one-form,
\begin{align*}
\nabla e_i=\sum_{k=1}^r A_{ki}\otimes e_k
\end{align*}
and
\begin{align*}
\nabla e_j=\sum_{k=1}^r A_{kj}\otimes e_k.
\end{align*}
Using that $h$ is linear in the first argument, conjugate-linear in the second, and $h(e_k,e_j)=\delta_{kj}$, we compute
\begin{align*}
h(\nabla e_i,e_j)=\sum_{k=1}^r A_{ki} h(e_k,e_j)=A_{ji}.
\end{align*}
Similarly,
\begin{align*}
h(e_i,\nabla e_j)=\sum_{k=1}^r \overline{A_{kj}} h(e_i,e_k)=\overline{A_{ij}}.
\end{align*}
Therefore
\begin{align*}
A_{ji}+\overline{A_{ij}}=0
\end{align*}
for all $i,j$. Replacing $(i,j)$ by $(j,i)$ gives
\begin{align*}
A_{ij}+\overline{A_{ji}}=0.
\end{align*}
Since $(A^*)_{ij}=\overline{A_{ji}}$, this is exactly the entrywise identity
\begin{align*}
A^*+A=0.
\end{align*}
[guided]
Assume that $\nabla$ is compatible with the Hermitian metric $h$. The point of using a unitary frame is that the metric coefficients of the frame are constant. For each pair of indices $i,j \in \{1,\dots,r\}$, the function
\begin{align*}
h(e_i,e_j):U \to \mathbb{C}
\end{align*}
is equal to the constant $\delta_{ij}$. Hence its [exterior derivative](/theorems/1525) is zero:
\begin{align*}
d(h(e_i,e_j))=0.
\end{align*}
Metric compatibility says that differentiating a Hermitian [inner product](/page/Inner%20Product) is the same as differentiating each section using the connection:
\begin{align*}
d(h(e_i,e_j))=h(\nabla e_i,e_j)+h(e_i,\nabla e_j).
\end{align*}
Since the left-hand side is zero, we get
\begin{align*}
0=h(\nabla e_i,e_j)+h(e_i,\nabla e_j).
\end{align*}
Now we insert the connection one-form. By definition of $A=(A_{ij})$, the $j$-th frame vector satisfies
\begin{align*}
\nabla e_j=\sum_{k=1}^r A_{kj}\otimes e_k.
\end{align*}
Thus, for the $i$-th frame vector,
\begin{align*}
\nabla e_i=\sum_{k=1}^r A_{ki}\otimes e_k.
\end{align*}
Because the Hermitian metric is linear in the first argument and the frame is unitary, we have
\begin{align*}
h(\nabla e_i,e_j)=\sum_{k=1}^r A_{ki}h(e_k,e_j)=\sum_{k=1}^r A_{ki}\delta_{kj}=A_{ji}.
\end{align*}
Because the Hermitian metric is conjugate-linear in the second argument, the one-form coefficients are conjugated in the second term:
\begin{align*}
h(e_i,\nabla e_j)=\sum_{k=1}^r \overline{A_{kj}}h(e_i,e_k)=\sum_{k=1}^r \overline{A_{kj}}\delta_{ik}=\overline{A_{ij}}.
\end{align*}
Substituting these two expressions into the compatibility identity gives
\begin{align*}
A_{ji}+\overline{A_{ij}}=0.
\end{align*}
Equivalently, after interchanging the names of the indices,
\begin{align*}
A_{ij}+\overline{A_{ji}}=0.
\end{align*}
But the $(i,j)$ entry of $A^*$ is $\overline{A_{ji}}$, so this entrywise identity is precisely
\begin{align*}
A^*+A=0.
\end{align*}
[/guided]
[/step]
[step:Recover metric compatibility for arbitrary sections from the skew-Hermitian identity]
Conversely, assume that
\begin{align*}
A^*+A=0.
\end{align*}
Equivalently, for all $i,j \in \{1,\dots,r\}$,
\begin{align*}
A_{ij}+\overline{A_{ji}}=0.
\end{align*}
Let $s,t \in \Gamma(U,E)$ be smooth local sections. Since $e$ is a local frame, there are unique smooth functions
\begin{align*}
s_i:U \to \mathbb{C}
\end{align*}
and
\begin{align*}
t_i:U \to \mathbb{C}
\end{align*}
for $1 \le i \le r$ such that
\begin{align*}
s=\sum_{i=1}^r s_i e_i
\end{align*}
and
\begin{align*}
t=\sum_{i=1}^r t_i e_i.
\end{align*}
Because the frame is unitary and $h$ is linear in the first argument,
\begin{align*}
h(s,t)=\sum_{i=1}^r s_i\overline{t_i}.
\end{align*}
Therefore the ordinary product rule gives
\begin{align*}
d(h(s,t))=\sum_{i=1}^r ds_i\,\overline{t_i}+\sum_{i=1}^r s_i\,d\overline{t_i}.
\end{align*}
Using the Leibniz rule for the connection and the definition of $A$,
\begin{align*}
\nabla s=\sum_{k=1}^r \left(ds_k+\sum_{i=1}^r A_{ki}s_i\right)\otimes e_k
\end{align*}
and
\begin{align*}
\nabla t=\sum_{k=1}^r \left(dt_k+\sum_{j=1}^r A_{kj}t_j\right)\otimes e_k.
\end{align*}
Pairing these with $t$ and $s$, respectively, gives
\begin{align*}
h(\nabla s,t)=\sum_{k=1}^r \left(ds_k+\sum_{i=1}^r A_{ki}s_i\right)\overline{t_k}
\end{align*}
and
\begin{align*}
h(s,\nabla t)=\sum_{k=1}^r s_k\left(d\overline{t_k}+\sum_{j=1}^r \overline{A_{kj}}\overline{t_j}\right).
\end{align*}
Adding,
\begin{align*}
h(\nabla s,t)+h(s,\nabla t)=\sum_{k=1}^r ds_k\,\overline{t_k}+\sum_{k=1}^r s_k\,d\overline{t_k}+\sum_{i,k=1}^r A_{ki}s_i\overline{t_k}+\sum_{k,j=1}^r s_k\overline{A_{kj}}\overline{t_j}.
\end{align*}
Rename the indices in the last sum by setting $i=k$ and $k=j$. Then the two connection-coefficient sums combine as
\begin{align*}
\sum_{i,k=1}^r \left(A_{ki}+\overline{A_{ik}}\right)s_i\overline{t_k}.
\end{align*}
The skew-Hermitian identity gives $A_{ki}+\overline{A_{ik}}=0$, so the connection-coefficient contribution vanishes. Hence
\begin{align*}
h(\nabla s,t)+h(s,\nabla t)=\sum_{k=1}^r ds_k\,\overline{t_k}+\sum_{k=1}^r s_k\,d\overline{t_k}=d(h(s,t)).
\end{align*}
Thus $\nabla$ is compatible with $h$ on $U$.
[/step]
