Lemma 11.20.1.B46ABA0FEC10 The curvature of $h$ is
\[ F_ h=\sum _ i\widehat{p_ i^*F_{h_ i}}. \]
For every $v\in E_ x$, the coefficient form of $\alpha _{h,v}$ is block diagonal with respect to the decomposition $T_ x^{1,0}X=\bigoplus _ iT_{x_ i}^{1,0}X_ i$, and
\[ \operatorname {tr}_{\omega _ i}\bigl (\alpha _{h,v}|_{T_{x_ i}X_ i}\bigr ) =h\bigl (\widehat K_ i v,v\bigr ),\qquad K_ i=K_{\omega _ i,h_ i}. \]
Proof.
In product holomorphic frames, the matrix of the metric is $M=\bigotimes _ iM_ i$, and its connection matrix is
\[ M^{-1}\partial M=\sum _ i\widehat{M_ i^{-1}\partial M_ i}. \]
Each summand depends only on the coordinates of $X_ i$. Applying $\bar\partial $ gives the curvature formula and shows that all components mixing different factors vanish. Pairing with $v$ gives the block diagonal form. Contracting the $i$th block with $\omega _ i$ commutes with the identity actions on the other tensor factors, which gives the trace identity.
$\square$
Pome
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