Kahler manifold computation In the following derivation, done on a Kahler manifold, where $\nabla$ is the complexification of the Riemannian connection (i.e., since we are on a Kahler manifold, this is the same thing as the Chern connection on holomorphic forms extended to the conjugate bundle in the natural way):

Where $\nabla_{i} \phi_{I_{p} \overline{J_{q}}} $ means $(\nabla_{i} \phi)_{I_{p} \overline{J_{q}}}$, i.e. we are taking a component of $\nabla_{i} \phi$, rather than taking the covariant derivative of a smooth function. I do not immediately see why it is clear that the symmetry of Christoffel symbols is enough to ensure the equality in the final line. It would be great if anyone could expand on why this is immediate.
Thank you in advance.
 A: This answer is a more coherent explanation of the comments I wrote to the question. First, I think it might be important to note that a version of this Proposition holds in less restrictive situations than Kähler manifolds.
Claim. Let $D$ be a torsion-free connection on $TM$. We can extend $D$ to a connection on $\Lambda^{k}T^*M$, using the duality between $TM$ and $T^*M$. Then, if $\phi$ is a $k$-form on $M$, the differential $\mathrm{d}\phi$ can be written as
\begin{equation}
\mathrm{d}\phi=D_a\phi_{b_1\dots b_n}\mathrm{d}x^a\wedge\mathrm{d}x^{b_1}\wedge\ldots\wedge\mathrm{d}x^{b_k}.
\end{equation}
Notice that this is not saying that $\mathrm{d}\phi=D\phi$, as the proof in the original question seems to imply. Instead, this claim tells us that the completely anti-symmetric part of $D\phi$ is just $\mathrm{d}\phi$, as a consequence of $D$ being torsion-free.
So, how do we prove this? As mentioned in the comments above, the crucial point is that the wedge product $\mathrm{d}x^a\wedge\mathrm{d}x^{b_1}\wedge\ldots\wedge\mathrm{d}x^{b_k}$ is completely anti-symmetric, i.e. anti-symmetric in all indices, while the Christoffel symbols $\Gamma^a_{bc}$ of $D$ are symmetric in $b,c$, as $D$ is torsion-free. So, we can compute
\begin{equation}
\begin{split}
\mathrm{d}\phi=&\partial_a\phi_{b_1\dots b_n}\mathrm{d}x^a\wedge\mathrm{d}x^{b_1}\wedge\ldots\wedge\mathrm{d}x^{b_k}=\\
=&D_a\phi_{b_1\dots b_n}\mathrm{d}x^a\wedge\mathrm{d}x^{b_1}\wedge\ldots\wedge\mathrm{d}x^{b_k}+\\
&+\sum_j\Gamma^{c}_{a b_j}\phi_{b_1\dots b_{j-1}c\,b_{j+1}\dots b_n}\mathrm{d}x^a\wedge\mathrm{d}x^{b_1}\wedge\ldots\wedge\mathrm{d}x^{b_k}=\\
=&D_a\phi_{b_1\dots b_n}\mathrm{d}x^a\wedge\mathrm{d}x^{b_1}\wedge\ldots\wedge\mathrm{d}x^{b_k}
\end{split}
\end{equation}
as $\Gamma^c_{ab_j}\mathrm{d}x^a\wedge\mathrm{d}x^{b_1}\wedge\ldots\wedge\mathrm{d}x^{b_k}=0$, since we are contracting a symmetric object with an anti-symmetric one.
