PomeProposition 11.2.8.E345C8B2A48C If $\mathrm d\theta =0$, then $DE_{\theta ,u}(v)=\int _ Xv\theta _ u^ n$ for every pair of smooth real functions $u,v$.
Proposition 11.2.8E345C8B2A48C
Proof. Differentiate the finite sum in (11.2.7). The term indexed by $j$ contributes
\[ \int _ Xv\theta _ u^ j\wedge \theta ^{n-j} +j\int _ Xu\mathrm{dd}^{\mathrm{c}}v\wedge \theta _ u^{j-1}\wedge \theta ^{n-j}, \]
the second summand being absent for $j=0$. For $j\geq 1$, the form $\theta _ u^{j-1}\wedge \theta ^{n-j}$ is closed, so by lemma 11.2.2 we may interchange $u$ and $v$ in the second summand. Using $\mathrm{dd}^{\mathrm{c}}u=\theta _ u-\theta $, we can then write the contribution of the term with $j\geq 1$ as
\[ (j+1)\int _ Xv\theta _ u^ j\wedge \theta ^{n-j} -j\int _ Xv\theta _ u^{j-1}\wedge \theta ^{n-j+1}. \]
Together with the contribution $\int _ Xv\theta ^ n$ of the term $j=0$, these expressions telescope to $(n+1)\int _ Xv\theta _ u^ n$. Dividing by $n+1$ completes the proof. $\square$
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