.\qedhere\end{aligned}$$

A straightforward (though lengthy) computation shows the following:

\[Lem:localstrAbdsH\] Suppose that $(z^1,\dots,z^n)$ is a local holomorphic chart and $H=\pi+\theta+\omega$ is given by $$\label{Eqt:Hlocally}
H= {\pi^{i,j}}{{\tfrac{\partial}{\partial{z^{i}}}}}\wedge{{\tfrac{\partial}{\partial{z^{j}}}}}
+{\theta^{p}_{q}}{{\tfrac{\partial}{\partial{z^{p}}}}}\wedge{d \bar{z}^{q}} 
+{\omega_{k,l}}{d \bar{z}^{k}}\wedge{d \bar{z}^{l}}
,$$ where ${\pi^{i,j}}$, ${\theta^{p}_{q}}$, and ${\omega_{k,l}}$ are complex valued smooth functions on ${\boldsymbol{X}}$. Then the $H$-twisted Lie algebroid structure on ${{A}^*}_H\cong{{T^{0,1}{\boldsymbol{X}}}}\oplus{{(T^{1,0}{\boldsymbol{X}})^*}}$ can be expressed by: $$\begin{gathered}
{{{a}_*}^H}({{\tfrac{\partial}{\partial{\bar{z}^{i}}}}})={{\tfrac{\partial}{\partial{\bar{z}^{i}}}}}-{\theta^{p}_{i}}{{\tfrac{\partial}{\partial{z^{p}}}}},
\qquad {{{a}_*}^H}({d z^{i}})=2{\pi^{i,q}}{{\tfrac{\partial}{\partial{z^{q}}}}}, \label{align:anchorsHlocal} \\
{[{{\tfrac{\partial}{\partial{\bar{z}^{i}}}}},{{\tfrac{\partial}{\partial{\bar{z}^{j}}}}}]^H_*}=2{\partial}{\omega_{i,j}},
\quad {[{d z^{i}},{d z^{j}}]^H_*}=2{\partial}{\pi^{i,j}},
\quad {[{d z^{j}},{{\tfrac{\partial}{\partial{\bar{z}^{i}}}}}]^H_*}={\partial}{\theta^{j}_{i}}. \label{align:basHlocal}\end{gathered}$$

Making the same assumptions as in Lemma \[Lem:localstrAbdsH\], consider the local section $$\label{Eqt:s} s={d z^{1}}\wedge\cdots\wedge{d z^{n}}$$ of ${\mathscr{L}}={Q^{{\tfrac{1}{2}}}_{{{A}^*}_H}}$. The representation of ${{A}^*}_H$ on ${\mathscr{L}}$ is given by $$\label{Eqt:RepHs}
{\nabla^H}_{{{\tfrac{\partial}{\partial{\bar{z}^{i}}}}}}s=-\tfrac{\partial{\theta^{p}_{i}}}{\partial z^p}s ,
\qquad {\nabla^H}_{{d z^{i}}}s=2\tfrac{\partial{\pi^{i,p}}}{\partial z^p}s
.$$

Using Eq. , we compute $$\label{eqt:ldanchorsHonAbd}
\begin{aligned}
{L_{{{{a}_*}^H}({{\tfrac{\partial}{\partial{\bar{z}^{i}}}}})}}{{d z^{j}}} &= -d{\theta^{j}_{i}}, &
{L_{{{{a}_*}^H}({{\tfrac{\partial}{\partial{\bar{z}^{i}}}}})}}{{d \bar{z}^{j}}} &= 0, \\
{L_{{{{a}_*}^H}({d z^{i}})}}{{d z^{j}}} &= 2 d{\pi^{i,j}}, &
{L_{{{{a}_*}^H}({d z^{i}})}}{{d \bar{z}^{j}}} &= 0.
\end{aligned}$$

Write $$s^2=({{\tfrac{\partial}{\partial{\bar{z}^{1}}}}}\wedge\cdots\wedge{{\tfrac{\partial}{\partial{\bar{z}^{n}}}}}\wedge{d z^{1}}\wedge\cdots\wedge{d z^{n}})
\otimes({d z^{1}}\wedge\cdots\wedge{d z^{n}}\wedge{d \bar{z}^{1}}\wedge\cdots\wedge {d \bar{z}^{n}}) .$$ Then, using Eqs.  and , one obtains $${\nabla^H}_{{{\tfrac{\partial}{\partial{\bar{z}^{i}}}}}}s^2=-2\tfrac{\partial{\theta^{p}_{i}}}{\partial z^p} s^2, 
\qquad {\nabla^H}_{{d z^{i}}}s^2=4\tfrac{\partial{\pi^{i,p}}}{\partial z^p}s^2 .$$ The conclusion thus follows immediately.

Locally, the operator ${{\mathcal{D}}^H}$ in Eq.  is given by $$\label{Eqt:DDHslocal}
{{\mathcal{D}}^H}s=(2\tfrac{\partial{\pi^{i,p}}}{\partial z^p}{{\tfrac{\partial}{\partial{z^{i}}}}} 
-\tfrac{\partial{\theta^{p}_{i}}}{\partial z^p}{d \bar{z}^{i}})\otimes s
,$$ where $s$ is defined in Eq. .

We are now ready to prove Theorem 