Lemma 14.13.2.6168D5D4EF96 At a point in a unitary frame,
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\begin{align} 2(n-1)Z_{i\bar j} ={}& \delta _{ij}\sum _{p,k}v_ p\overline{T_{pk\bar k}} -v_ i\sum _ k\overline{T_{jk\bar k}} -\sum _ kv_ k\overline{T_{kj\bar i}}\notag \\ & +\delta _{ij}\sum _{p,k}v_{\bar p}T_{pk\bar k} -v_{\bar j}\sum _ kT_{ik\bar k} -\sum _ kv_{\bar k}T_{ki\bar j}. \tag{14.13.3} \end{align}
In particular $Z_{i\bar j}$ contains no $v_ j$ or $v_{\bar i}$. Moreover
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14.13.4
\begin{equation} 2(n-1)Z_{i\bar i} =\sum _{p\ne i,\ k\ne i} \bigl (v_ p\overline{T_{pk\bar k}}+v_{\bar p}T_{pk\bar k}\bigr ). \end{equation}
Covariant differentiation preserves these coefficient identities. For example $\nabla _1 Z_{1\bar1}$ contains no pure Hessian $v_{11}$ or its conjugate, and $\nabla _{\bar1}\nabla _1 Z_{1\bar1}$ contains no $v_{11\bar1}$ or its conjugate, nor a pure $v_{11}$ or its conjugate.
Proof.
For $n=2$, the defining expression of $Z$ is zero, and so are the formulas below, because the generalized Kronecker symbol with three indices vanishes in two dimensions. Suppose $n\geq 3$ and use coordinates that are unitary at the point. Direct antisymmetrization gives
\[ \bar\partial \gamma =-\frac{\mathrm i}{2}\sum _{a,b,c} \overline{T_{bc\bar a}}\, \mathrm{d}z^ a\wedge \mathrm{d}\bar z^ b\wedge \mathrm{d}\bar z^ c. \]
We use $\bar\partial \gamma ^{n-2}=(n-2)\bar\partial \gamma \wedge \gamma ^{n-3}$. To write the contraction explicitly, put
\[ \delta ^{paj}_{bci} =\det \begin{pmatrix} \delta _{pb}
& \delta _{pc}
& \delta _{pi}
\\ \delta _{ab}
& \delta _{ac}
& \delta _{ai}
\\ \delta _{jb}
& \delta _{jc}
& \delta _{ji}
\end{pmatrix}. \]
Let $A=\star _\gamma (\mathrm i\partial v\wedge \bar\partial \gamma ^{n-2})/(n-1)!$. Expanding the $n-3$ remaining copies of $\gamma $ leaves the three complementary indices and gives
\[ A_{i\bar j}=\frac1{2(n-1)} \sum _{p,a,b,c}v_ p\overline{T_{bc\bar a}}\, \delta ^{paj}_{bci}. \]
Indeed, the factor is $(n-2)(n-3)!/[2(n-1)!]$, and the determinant records the signs of the remaining holomorphic and antiholomorphic indices. Expanding its six terms and using $T_{bc\bar a}=-T_{cb\bar a}$, we obtain
\[ A_{i\bar j}=\frac1{n-1}\left( \delta _{ij}\sum _{p,k}v_ p\overline{T_{pk\bar k}} -v_ i\sum _ k\overline{T_{jk\bar k}} -\sum _ kv_ k\overline{T_{kj\bar i}}\right). \]
Since $Z=(A+A^*)/2$, where $A^*$ denotes the Hermitian adjoint matrix, this gives (14.13.3).
For $i\ne j$, the coefficient of $v_ j$ in this expression vanishes because $T_{jj\bar i}=0$, and the coefficient of $v_{\bar i}$ vanishes in the conjugate expression. For $i=j$, canceling the terms with $p=i$ and $k=i$ gives (14.13.4). These are tensor identities valid in every unitary frame. Since the Chern connection preserves the metric, they can be differentiated in a unitary frame that is parallel at the point. For instance, the first derivative of a summand in (14.13.4) is
\[ (\nabla _1v_ p)\overline{T_{pk\bar k}} +v_ p\nabla _1\overline{T_{pk\bar k}},\qquad p,k\ne 1, \]
together with its conjugate counterpart. A further derivative in the direction $\bar1$ still has $p\ne 1$. This gives the stated exclusions, including the terms in which a derivative falls on a coefficient.
$\square$
Pome
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