PomeLemma 11.9.3.EE774D339C2C With $R_ j$ the curvature of $k_ j$,
\[ \beta _ s(G)=\mathrm{dd}^{\mathrm{c}}U_ s-\frac12\sum _ jR_ jV_{sj}. \]
Lemma 11.9.3EE774D339C2C
Proof. At a fixed base point, choose holomorphic frames of the line bundles with $k_ j=\mathrm e^{-\varphi _ j}$ and $\mathrm d\varphi _ j=0$ at that point. There the horizontal lift has zero vertical component. Write the smooth unitary fiber coordinates as $\zeta _ j=\mathrm e^{-\varphi _ j/2}v_ j$ and $G(z,v)=g(z,\zeta )$. At the base point, the first base derivatives of $\zeta _ j$ vanish, and its mixed second base derivatives are $-\frac12(\varphi _ j)_{\alpha \bar\beta }\zeta _ j$ and their conjugates. The chain rule therefore gives
\[ G_{\alpha \bar\beta } =g_{\alpha \bar\beta }-\frac12\sum _ j(\varphi _ j)_{\alpha \bar\beta } (\zeta _ jg_{\zeta _ j}+\bar\zeta _ jg_{\bar\zeta _ j}). \]
The expression in parentheses is $D_ jG$. Both the spherical measure and $\rho _ s$ are independent of $z$ in unitary coordinates, so averaging the identity gives the result, since $R_ j=\mathrm{dd}^{\mathrm{c}}\varphi _ j$. $\square$
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