Proof. Let \(J_Y\) be the complex structure on \(Y\). On each chart \(f^{-1}(U_a)\), we define \(J_a|_{X_s} = \Phi_{a,s}^* J_Y\). On the overlap \(U_{ab}\), we have \(\Phi_{a,s} = g_{ab}(s) \comp \Phi_{b,s}\). Thus, \[J_a|_{X_s} = (g_{ab}(s) \comp \Phi_{b,s})^* J_Y = \Phi_{b,s}^* g_{ab}(s)^* J_Y=\Phi_{b,s}^* J_Y=J_b|_{X_s}.\] These local structures glue to a global fiberwise complex structure \(J_{X/S}\). ◻

Proof. Let \(Q = TX^{\CC}/ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup\). Then we have the natural projection \(p: TX^{\CC} \to Q\), such that \(\mathrm{d}f^{\CC} = q\comp p\). Let \(TX\) be an almost complex structure on \((f, T_{X/S})\). Consider the restriction \(\bar{\pi} = \mathrm{d}f^{\CC}|_{ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup }: \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup \to f^* \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TS} \endgroup\). Its kernel is \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup \cap T_{X/S}^{\CC}\). We claim this intersection is \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup\). Clearly \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup\) is contained in the intersection. Conversely, let \(v \in \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup \cap T_{X/S}^{\CC}\). We decompose \(v=v^{1,0}+v^{0,1}\) with \(v^{1,0}\in T_{X/S}, v^{0,1}\in \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup\). Since \(v^{0,1}\in \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup \subset \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\) and \(v\in \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\), we have \(v^{1,0}=v-v^{0,1} \in \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\). But \(v^{1,0}\in T_{X/S}\subset TX\). Thus \(v^{1,0}\in TX\cap \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup =\{0\}\). So \(v\in \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup\). Therefore, we have a short exact sequence \[\label{eq:barTX95SES} 0 \to \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup \to \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup \xrightarrow{\bar{\pi}} f^* \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TS} \endgroup \to 0.\tag{5}\] For \(u\in f^* \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TS} \endgroup\), let \(w\in \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\) be any lift such that \(\bar{\pi}(w)=u\). Define \(\bar{\partial}(u) := p(w)\). This is well-defined because if \(w'\) is another lift, \(w-w'\in \ker(\bar{\pi})= \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup =\ker(p)\). It satisfies the definition of a \(\bar{\partial}\)-operator since \(q(\bar{\partial}(u)) = q(p(w)) = \mathrm{d}f^{\CC}(w) = u\).

Conversely, given a \(\bar{\partial}\)-operator, we define \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup = p^{-1}(\im(\bar{\partial}))\). Let \(TX = \overline{ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup }\). \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup =\ker(p) \subset \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\). By conjugation, \(T_{X/S}\subset TX\). \(\mathrm{d}f^{\CC}( \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup ) = q(p( \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup )) = q(\im(\bar{\partial}))\). Since \(\bar{\partial}\) is a splitting over \(f^* \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TS} \endgroup\), this is \(f^* \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TS} \endgroup\). By conjugation, \(\mathrm{d}f^{\CC}(TX)\subset f^*TS\). Let \(v\in TX\cap \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\), \(\mathrm{d}f^{\CC}(v) \in f^*TS \cap f^* \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TS} \endgroup = \{0\}\), so \(v\in T_{X/S}^{\CC}\). Suppose \(w\in \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup \cap T_{X/S}^{\CC}\). The condition \(w\in T_{X/S}^{\CC}\) implies \(\mathrm{d}f^{\CC}(w)=0\), while \(w\in \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\) means \(p(w)\in \im(\bar{\partial})\). Thus \(q(p(w))=\mathrm{d}f^{\CC}(w)=0\). Since \(q|_{\im(\bar{\partial})}\) is an isomorphism onto \(f^* \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TS} \endgroup\), we must have \(p(w)=0\). Thus \(w\in \ker(p)= \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup\). So \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup \cap T_{X/S}^{\CC} = \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup\). Therefore, \(v \in T_{X/S}^{\CC} \cap TX \cap \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup = T_{X/S} \cap \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup = \{0\}\). By rank reasons, \(TX^{\CC}=TX\oplus \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\).

These constructions are mutually inverse. The final statement regarding integrability follows, as the integrability of \(TX\) is precisely the definition of the integrability of the corresponding \(\bar{\partial}\). ◻

Proof. By the definition of \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\), we have the short exact sequence 5 . A lifting of \(\bar{\partial}\) is a splitting of the projection \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup \to f^* \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TS} \endgroup\). Write \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup = \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup \oplus \im(\nabla^{0,1})\). By the lifting condition, \([ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup , \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup ]\subset \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\) holds if and only if \([\im(\nabla^{0,1}),\textrm{im}(\nabla^{0,1})]\subset \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\) holds. Equivalently, the composite \[f^*\wedge^2 \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TS} \endgroup \xrightarrow{\frac{1}{2}[\nabla^{0,1},\nabla^{0,1}]} TX^\CC\to \frac{TX^\CC}{ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup }, \quad a\wedge b \mapsto [[\nabla^{0,1}(a),\nabla^{0,1}(b)]],\] is zero. Suppose the \(\bar{\partial}\)-operator is integrable. By Proposition 9, this induces a complex structure on \(X\) such that \(f: X\to S\) is a holomorphic fibration. Any smooth splitting \(\nabla^{0,1}\) of \[0\to \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup \to \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup \to f^\ast \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TS} \endgroup \to 0\] is a lifting of \(\bar{\partial}\). Moreover, \([ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup ,\im(\nabla^{0,1})]\subset [ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup , \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup ]\subset \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\). The lifting condition is satisfied. ◻

Proof. The block triangular form follows from the compatibility of \(J\) with the complex fiber bundle structure. Let \(v \in f^* TS^\RR\) and let \(H_{v^{0,1}}\) be the horizontal lift of \(v^{0,1}\) with respect to \(\nabla^{0,1}\). By the definition of \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup\), we have \[J(H_{v^{0,1}}- \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup (v^{0,1}))=-\mathrm{i}(H_{v^{0,1}}- \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup (v^{0,1})).\] Since \(J( \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup (v^{0,1}))=\mathrm{i} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup (v^{0,1})\), we have \(J(H_{v^{0,1}}) = -\mathrm{i}H_{v^{0,1}} + 2\mathrm{i} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup (v^{0,1})\). The real horizontal lift decomposes as \(H_v=H_{v^{0,1}}+ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{H_{v^{0,1}}} \endgroup\). Since \(J\) is a real operator, \[J(H_v) = J(H_{v^{0,1}}) + \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{J(H_{v^{0,1}})} \endgroup = \bigl(-\mathrm{i}H_{v^{0,1}} + 2\mathrm{i} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup (v^{0,1})\bigr) + \bigl(\mathrm{i} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{H_{v^{0,1}}} \endgroup - 2\mathrm{i}\overline{ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup (v^{0,1})}\bigr).\] Therefore, \(\phi(v) =2\mathrm{i} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup (v^{0,1}) + \overline{2\mathrm{i} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup (v^{0,1})} = 4 \Re(\mathrm{i} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup (v^{0,1}))\). ◻

Proof. \(\bar{\partial}\) satisfies the lifting condition if and only if \([ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup ,\im(\nabla^{0,1})]\subset \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\). We compute \[[\partial_{\bar{\beta}}, H_{\bar{j}}] = [\partial_{\bar{\beta}}, \partial_{\bar{j}}+\Gamma^{\gamma}_{\bar{j}}\partial_{\gamma}+\Gamma^{\bar{\delta}}_{\bar{j}}\partial_{\bar{\delta}}] = (\partial_{\bar{\beta}}\Gamma^{\gamma}_{\bar{j}})\partial_{\gamma} + (\partial_{\bar{\beta}}\Gamma^{\bar{\delta}}_{\bar{j}})\partial_{\bar{\delta}}.\] This vector belongs to \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{TX} \endgroup\) if and only if \(\partial_{\bar{\beta}}\Gamma_{\bar{j}}^\gamma=0\).

To show the condition 14 is well-defined, we check the transformation law using 11 . \[\begin{align} \partial_{\bar{\beta}'} {\Gamma'}_{\bar{j}}^{\gamma} &= \frac{\partial \bar{s}^k}{\partial \bar{s}'^j} \partial_{\bar{\beta}'} \Bigl( \frac{\partial z'^\gamma}{\partial \bar{s}^k} + \frac{\partial z'^\gamma}{\partial z^\alpha} \Gamma_{\bar{k}}^\alpha \Bigr)\\ &= \frac{\partial \bar{s}^k}{\partial \bar{s}'^j} \Bigl( 0 + \frac{\partial z'^\gamma}{\partial z^\alpha} (\partial_{\bar{\beta}'} \Gamma_{\bar{k}}^\alpha) + 0 \Bigr)\\ &=\frac{\partial \bar{s}^k}{\partial \bar{s}'^j} \frac{\partial z'^\gamma}{\partial z^\alpha} \frac{\partial \bar{z}^\delta}{\partial \bar{z}'^\beta} (\partial_{\bar{\delta}} \Gamma_{\bar{k}}^\alpha)=0. \end{align}\] Hence 14 is well-defined. ◻

Proof. Fix \(\bar{\partial}_0 \in \mathcal{A}^{0,1}_{\mathrm{LC}}\). By Lemma 2, any other \(\bar{\partial}\)-operator is of the form \(\bar{\partial}= \bar{\partial}_0 + \Phi\), where \(\Phi \in C^{\infty}(X,f^\ast \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T^* S} \endgroup \otimes T_{X/S})\). Let \(\Gamma_{0,\bar{j}}^\gamma\) be the local coefficients of \(\bar{\partial}_0\), and \(\Phi_{\bar{j}}^\gamma\) be the coefficients of \(\Phi\). The coefficients of \(\bar{\partial}\) are \(\Gamma_{\bar{j}}^\gamma = \Gamma_{0,\bar{j}}^\gamma + \Phi_{\bar{j}}^\gamma\). By Lemma 5, we have \(\partial_{\bar{\beta}}\Gamma_{0,\bar{j}}^\gamma=0\). Then \(\bar{\partial}\) satisfies the lifting condition if and only if \(\partial_{\bar{\beta}}\Gamma_{\bar{j}}^\gamma=0\), which is equivalent to \(\partial_{\bar{\beta}}\Phi_{\bar{j}}^\gamma=0\). This condition means that \(\Phi\) is holomorphic along the fibers, and can be identified with an element in \(A^{0,1}(S,f_\ast T_{X/S})\). ◻

Proof. We verify the transformation law for \(\nabla_1^{1,0}\). We have \({\Gamma_1}_i^\alpha=\Gamma_i^\alpha\) and \({\Gamma_1}_i^{\bar \beta}=\overline{\Gamma_{\bar i}^\beta}\). Then by 10 and 11 , \[\begin{align} {\Gamma_1'}_i^\alpha&=\frac{\partial s^j}{\partial s'^i}(\partial_j+\Gamma_j^\beta \partial_\beta)z'^\alpha=\frac{\partial s^j}{\partial s'^i}(\partial_j+{\Gamma_1}_j^\beta \partial_\beta)z'^\alpha,\\ {\Gamma_1'}_i^{\bar \beta}&=\overline{\frac{\partial \bar{s}^j}{\partial \bar{s}'^i}(\partial_{\bar{j}}+\Gamma_{\bar{j}}^\alpha \partial_\alpha)z'^\beta}=\frac{\partial s^j}{\partial s'^i}(\partial_j+{\Gamma_1}_j^{\bar{\alpha}}\partial_{\bar{\alpha}}) \bar{z}'^{\beta}. \end{align}\] The connection \(\nabla_1^{1,0}\) is globally well-defined. ◻

Proof. First we show that the condition 20 is well-defined. In fact, by 10 , \[\begin{align} \partial_{\bar{\beta}'} {\Gamma'}_{i}^{\alpha} &= \partial_{\bar{\beta}'} \Big( \frac{\partial s^j}{\partial s'^i} \left( \partial_j z'^\alpha + \Gamma_j^\gamma \partial_\gamma z'^\alpha \right) \Big)=\frac{\partial s^j}{\partial s'^i} \partial_{\bar{\beta}'} \left( \partial_j z'^\alpha + \Gamma_j^\gamma \partial_\gamma z'^\alpha \right)\\ &=\frac{\partial s^j}{\partial s'^i} \big(\partial_{\bar{\beta}'}(\partial_j z'^\alpha) + (\partial_{\bar{\beta}'} \Gamma_j^\gamma) \partial_\gamma z'^\alpha + \Gamma_j^\gamma \partial_{\bar{\beta}'} (\partial_\gamma z'^\alpha) \big)\\ &=\frac{\partial s^j}{\partial s'^i}(\partial_{\bar{\beta}'} \Gamma_j^\gamma) \partial_\gamma z'^\alpha=0. \end{align}\]

For \(H_i=\nabla^{1,0}\partial_i=\partial_i+\Gamma_i^\alpha\partial_\alpha+\Gamma_i^{\bar{\beta}}\partial_{\bar{\beta}}\in\im(\nabla^{1,0})\) and \(\partial_{\bar{\gamma}} \in \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup\), we have \[[\partial_{\bar{\gamma}}, H_i] = (\partial_{\bar{\gamma}}\Gamma_i^\alpha) \partial_\alpha+(\partial_{\bar{\gamma}}\Gamma_i^{\bar{\beta}}) \partial_{\bar{\beta}}.\] So \(F^{1,1}_{D}\) is well-defined if and only if all \(\partial_{\bar{\gamma}}\Gamma_i^\alpha\) vanish. 21 follows from \[[\partial_i+\Gamma_i^\alpha\partial_\alpha+\Gamma_i^{\bar{\beta}}\partial_{\bar{\beta}}, \partial_{\bar{j}}+\Gamma_{\bar{j}}^\gamma\partial_\gamma]=\bigl( \partial_i \Gamma_{\bar{j}}^\alpha - \partial_{\bar{j}} \Gamma_i^\alpha + \Gamma_i^\beta \partial_\beta \Gamma_{\bar{j}}^\alpha - \Gamma_{\bar{j}}^\gamma \partial_\gamma \Gamma_i^\alpha + \Gamma_i^{\bar{\beta}} \partial_{\bar{\beta}} \Gamma_{\bar{j}}^\alpha \bigr) \partial_\alpha\mod \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup .\qedhere\] ◻

Proof. Let \(\nabla_1^{1,0},\nabla_2^{1,0}\in \mathcal{A}_{\mathrm{RH},\bar{\partial}}\). In adapted local coordinates, we write \[\nabla_1^{1,0}(\partial_i)=\partial_i+{\Gamma_1}_i^\alpha\partial_\alpha+{\Gamma_1}_i^{\bar{\beta}}\partial_{\bar{\beta}},\quad \nabla_2^{1,0}(\partial_i)=\partial_i+{\Gamma_2}_i^\alpha\partial_\alpha+{\Gamma_2}_i^{\bar{\beta}}\partial_{\bar{\beta}}.\] Since their conjugates induce the same \(\bar{\partial}\), we have \({\Gamma_1}_i^{\bar{\beta}}={\Gamma_2}_i^{\bar{\beta}}\). Then \[\label{TransgressionForm95eq} G_{1,2}:=\nabla_1^{1,0}-\nabla_2^{1,0}=({\Gamma_1}_i^\alpha-{\Gamma_2}_i^\alpha) \mathrm{d}s^i\otimes\partial_\alpha.\tag{23}\] We will prove that \(G_{1,2}\) is a global section of \(C^\infty(X, f^* T^*S \otimes T_{X/S})\) by showing it transforms correctly under coordinate changes \(s' = s'(s), z' = z'(s,\bar{s},z)\). By 10 , \[\begin{align} ({\Gamma'_1}_i^\alpha - {\Gamma'_2}_i^\alpha) \mathrm{d}{s'}^i\otimes \partial_{\alpha'} &= \frac{\partial s^j}{\partial s'^i} \big( \big( \partial_j z'^\alpha + {\Gamma_1}_j^\beta \partial_\beta z'^\alpha \big) - \big( \partial_j z'^\alpha + {\Gamma_2}_j^\beta \partial_\beta z'^\alpha \big) \big)\mathrm{d}{s'}^i\otimes\partial_{\alpha'} \\ &= \frac{\partial s^j}{\partial s'^i} \big( {\Gamma_1}_j^\beta - {\Gamma_2}_j^\beta \big) \partial_\beta z'^\alpha \mathrm{d}{s'}^i\otimes \partial_{\alpha'} \\ &= \frac{\partial s^j}{\partial s'^i} \big({\Gamma_1}_j^\beta - {\Gamma_2}_j^\beta\big) \frac{\partial s'^i}{\partial s^p} \mathrm{d}s^p\otimes \partial_\beta\\ &= ({\Gamma_1}_p^\beta - {\Gamma_2}_p^\beta) \mathrm{d}s^p\otimes \partial_\beta. \end{align}\] Therefore \(G_{1,2}\) is globally defined. Since \(\nabla_1^{1,0},\nabla_2^{1,0}\) are both relatively holomorphic, we have \(G_{1,2}\in A^{1,0}(S,f_*T_{X/S})\). On the other hand, for any \(\nabla_1^{1,0}\) and \(G_{1,2}\in A^{1,0}(S,f_*T_{X/S})\), we have \(\nabla_1^{1,0}+G_{1,2}\in \mathcal{A}_{\mathrm{RH},\bar{\partial}}\).

The map \(\mathcal{A}_{\mathrm{RH},\bar{\partial}}\to \mathcal{A}_{\mathrm{LC}}^{1,0}\) is provided by 22 . Its inverse is provided by Lemma 7, which determines \(\nabla^{1,0}\) using \(\partial\) and \(\bar{\partial}\). ◻

Proof. The antiholomorphic derivative transforms as \[\partial_{\bar{j}'} = \frac{\partial}{\partial \bar{s}'^j} = \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{ \frac{\partial s^k}{\partial s'^j}} \endgroup \partial_{\bar{k}}- \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{ \frac{\partial s^k}{\partial s'^j}} \endgroup \frac{\partial \bar{z}'^\beta}{\partial \bar{s}^k} \frac{\partial \bar{z}^\gamma}{\partial \bar{z}'^\beta}\partial_{\bar{\gamma}}.\] Apply \(\partial_{\bar{j}'}\) to the transformation formula 10 and use 20 , \[\begin{align} \partial_{\bar{j}'} \Gamma'{}_i^\alpha &= \partial_{\bar{j}'} \Big( \frac{\partial s^j}{\partial s'^i} \big( \partial_j z'^\alpha + \Gamma_j^\beta \partial_\beta z'^\alpha \big) \Big) \\ &= \frac{\partial s^j}{\partial s'^i} \partial_{\bar{j}'} \big( \Gamma_j^\beta \partial_\beta z'^\alpha \big) = \frac{\partial s^j}{\partial s'^i} \big( \partial_{\bar{j}'} \Gamma_j^\beta \big) \partial_\beta z'^\alpha \\ &= \frac{\partial s^j}{\partial s'^i} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{ \frac{\partial s^k}{\partial s'^j} } \endgroup \big( \partial_{\bar{k}} \Gamma_j^\beta \big) \partial_\beta z'^\alpha. \end{align}\] The relative tangent vectors transform as \[\partial_{\alpha'} = \frac{\partial z^\gamma}{\partial z'^\alpha} \partial_\gamma, \quad \text{thus} \quad \partial_\beta z'^\alpha \partial_{\alpha'} = \partial_\beta z'^\alpha \frac{\partial z^\gamma}{\partial z'^\alpha} \partial_\gamma = \delta_\beta^\gamma \partial_\gamma = \partial_\beta.\] Therefore \[\big( \partial_{\bar{j}'} \Gamma'{}_i^\alpha \big) \partial_{\alpha'} = \frac{\partial s^j}{\partial s'^i} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{ \frac{\partial s^k}{\partial s'^j} } \endgroup \big( \partial_{\bar{k}} \Gamma_j^\beta \big) \partial_\beta.\] The differential form transforms as \[\mathrm{d}s'^i \wedge \mathrm{d}\bar{s}'^j = \frac{\partial s'^i}{\partial s^p} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{ \frac{\partial s'^j}{\partial s^q} } \endgroup \mathrm{d}s^p \wedge \mathrm{d}\bar{s}^q.\] Then we have \[\begin{align} \big(\partial_{\bar{j}'} \Gamma'{}_i^\alpha\big) \mathrm{d}s'^i \wedge \mathrm{d}\bar{s}'^j\otimes \partial_{\alpha'} &= \Big( \frac{\partial s'^i}{\partial s^p} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{ \frac{\partial s'^j}{\partial s^q} } \endgroup \mathrm{d}s^p \wedge \mathrm{d}\bar{s}^q \Big)\otimes \Big( \frac{\partial s^j}{\partial s'^i} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{ \frac{\partial s^k}{\partial s'^j} } \endgroup \big( \partial_{\bar{k}} \Gamma_j^\beta \big) \partial_\beta \Big) \\ &= \frac{\partial s^j}{\partial s'^i} \frac{\partial s'^i}{\partial s^p} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{ \frac{\partial s^k}{\partial s'^j} \frac{\partial s'^j}{\partial s^q} } \endgroup \big( \partial_{\bar{k}} \Gamma_j^\beta \big) \mathrm{d}s^p \wedge \mathrm{d}\bar{s}^q\otimes \partial_\beta\\ &=\delta_p^j \delta_q^k\big( \partial_{\bar{k}} \Gamma_j^\beta \big) \mathrm{d}s^p \wedge \mathrm{d}\bar{s}^q\otimes \partial_\beta\\ &=\big( \partial_{\bar{q}} \Gamma_p^\beta \big) \mathrm{d}s^p \wedge \mathrm{d}\bar{s}^q\otimes \partial_\beta. \qedhere \end{align}\] ◻

Proof. Let \(\nabla_1^{1,0},\nabla_2^{1,0}\) be two relatively holomorphic pure \((1,0)\)-connections and \(F_1,F_2\) be the corresponding \((1,1)\)-type curvatures. By 25 , locally we have \(F_1-F_2=\bar{\partial}_X G_{1,2}\), where \(G_{1,2}\) is given by 23 , and \(G_{1,2}\in A^{1,0}(S,f_\ast T_{X/S})\subset C^{\infty}(X,f^*T^*S\otimes T_{X/S})\). Therefore \(F_1-F_2\) is \(\bar{\partial}_X\)-exact, and \([F_1]=[F_2]\) in \(H^1(X,f^*\Omega_S \otimes T_{X/S})\). ◻

Proof. Consider the Čech-Dolbeault double complex \(K^{p,q} = \check{C}^p(\mathcal{U}, \mathcal{A}^{0,q}(TX_{s_0}))\). By the differential geometric definition, the Kodaira-Spencer class \(\rho(\partial_i)\) is represented by a Čech 1-cocycle \(\theta \in \check{C}^1(\mathcal{U}, TX_{s_0})\), given by \(\theta_{ab} := (\partial_i f_{ab}^\alpha(s,\bar{s}, z_b) )\partial_{\alpha,a}\) on the overlap \(U_{ab}|_{X_{s_0}}\). By 10 , \[\begin{align} \theta_{ab}&=(\partial_i f_{ab}^\alpha) \partial_{\alpha,a}=\Gamma_{i,a}^\alpha\partial_{\alpha,a}-\Gamma_{i,b}^\beta(\partial_{\beta,b}f_{ab}^\alpha) \partial_{\alpha,a} \\ &=\Gamma_{i,a}^\alpha\partial_{\alpha,a}-\Gamma_{i,b}^\alpha\partial_{\alpha,b}=:\sigma_a-\sigma_b. \end{align}\] Therefore, we find a 0-cochain \(\sigma = \{\sigma_a\} \in K^{0,0}\) such that \(\delta\sigma=\theta\). On the other hand, \[\bar{\partial}\sigma_a=(\bar{\partial}_{X_{s_0}} \Gamma_{i,a}^\alpha)\partial_{\alpha,a}= (\partial_{\bar{\beta},a} \Gamma_{i,a}^\alpha) \mathrm{d}\bar{z}_a^\beta\otimes \partial_{\alpha,a}.\] This implies \(\bar{\partial}\sigma=b_i\), and the Kodaira-Spencer class \(\rho(\partial_i)\) is \(b_i\). The proof for \(\rho_{s_0}(\partial_{\bar i})\) is similar. ◻

Proof. Choose a compatible atlas \(\mathcal{U} = \{(U_a, \Phi_a)\}\) with trivializations \(\Phi_a: f^{-1}(U_a) \xrightarrow{\cong} U_a \times Y\). On each chart \(U_a\), the trivialization induces a canonical flat connection compatible with the product structure. We define the local horizontal lifts of \(\partial_i\) and \(\partial_{\bar{i}}\) as \(H_{i,a} := \Phi_a^* ( \partial_i)\), \(H_{\bar{i},a} := \Phi_a^* ( \partial_{\bar i})\). Let \(\{\chi_a\}\) be a smooth partition of unity on \(S\) subordinate to the cover \(\{U_a\}\). We define the global horizontal lifts by \[H_i := \sum_a f^*(\chi_a) H_{i,a}, \quad H_{\bar{i}} := \sum_a f^*(\chi_a) H_{\bar{i},a}.\] These define the connections \(\nabla^{1,0}\) and \(\nabla^{0,1}\).

We verify that \(\nabla^{1,0}\) is relatively holomorphic. It suffices to check this in an arbitrary chart \((U_b, \Phi_b)\) with coordinates \((s, z_b)\). The term \(H_{i,b}\) is just \(\partial_i\). Consider the term \(H_{i,a}\) from a different chart. The coordinate transformation is \(z_a = g_{ab}(s,\bar{s},z_b)\). The vector field \(H_{i,a}\) expressed in the \(b\)-coordinates is \[H_{i,a} =\partial_i + (\partial_i g_{ba}^\alpha(s,\bar{s},z_a)) \partial_{z_b^\alpha} + (\partial_i \bar{g}_{ba}^\beta(s,\bar{s}, z_a)) \partial_{\bar{z}_b^\beta}.\] Note that \(\partial_i g_{ba}^\alpha(s,\bar{s},z_a)\) is holomorphic in \(z_b\). The global connection coefficient is a convex combination \[\Gamma_i^\alpha(s, z_b) = \sum_a \chi_a(s) (\partial_i g_{ba}^\alpha(s,\bar{s}, z_a)),\] which is holomorphic in \(z_b\). The proof for \(\nabla^{0,1}\) is identical.

Now suppose that \(f\) has compact fibers and admits a relatively holomorphic \((1,0)\)-connection \(\nabla^{1,0}\) and a mixed relatively holomorphic \((0,1)\)-connection \(\nabla^{0,1}\). Using Lemma 7, we obtain a \((1,0)\)-connection \(\nabla_1^{1,0}\) from \(\nabla^{1,0}\) and \(\nabla^{0,1}\). \(\nabla_1^{1,0}\) is relatively holomorphic and \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\nabla_1^{1,0}} \endgroup\) is mixed relatively holomorphic, so by Lemma 20 below, the parallel transport maps defined by \(\nabla_1^{1,0}\) are biholomorphisms. Then \(f\) admits a compatible atlas provided by radial parallel transports as in Lemma 21, which means \(f\) is an isotrivial complex fiber bundle. If moreover \(f\) is a holomorphic fibration, since its fibers are compact and biholomorphic, \(f\) is a holomorphic fiber bundle by the Fischer–Grauert theorem. ◻

Proof. Note that locally we have \[\mathcal{R}=\bar{\partial}_X(\Gamma_i^\alpha \mathrm{d}s^i\otimes \partial_\alpha),\] so \(\mathcal{R}\) is \(\bar{\partial}_X\)-closed. By 23 , \([\mathcal{R}]\) is independent of \(\nabla\). The last statement follows from the above computations. ◻

Proof. A smooth section \(A\) of \(\ad P\) over \(U\subset S\) corresponds to a smooth map \(A:\pi_P^{-1}(U)\to\mathfrak{g}\) such that \(A(p\cdot g)=\Ad_{g^{-1}}A(p)\). We define a vertical vector field \(\widetilde{V}_A\) on \((\pi_P^{-1}(U))\times Y\) by \[\widetilde{V}_A\big|_{(p,y)} := \big(0,\tau_0(A(p))|_y\big).\] \(\widetilde{V}_A\) is invariant under the action \(r_g(p,y)=(p\cdot g,\,g^{-1}\!\cdot y)\) since \[\begin{align} (\mathrm{d}r_g)_{(p,y)}\bigl(\widetilde{V}_A\big|_{(p,y)}\bigr) &= \bigl(0,(\mathrm{d}L_{g^{-1}})_y\bigl(\tau_0(A(p))\bigr)\bigr)\\ &= \big(0,\tau_0\big(\Ad_{g^{-1}}A(p)\big)\big|_{g^{-1}\cdot y}\big) \quad (\text{by equivariance of } \tau_0)\\ &= \big(0,\tau_0\big(A(p\cdot g)\big)\big|_{g^{-1}\cdot y}\big) = \widetilde{V}_A\big|_{r_g(p,y)}. \end{align}\] Thus \(\widetilde{V}_A\) descends to a smooth vertical vector field \(V_A\) on \(X|_U\), which is fiberwise holomorphic. The assignment \(A\mapsto V_A\) defines a morphism of sheaves \(\tau:\ad P\to f_*T_{X/S}\), which is injective as \(\tau_0\) is.

If \(\tau_0\) is an isomorphism, then \(\dim_\CC H^0(Y, TY)=\dim_\CC \mathfrak{g}\), which is finite. The above construction provides a bijection between equivariant maps into \(\mathfrak{g}\) and equivariant maps into \(H^0(Y, TY)\), hence \(\tau\) is an isomorphism of complex vector bundles. ◻

Proof. Consider the local smooth trivialization \(P|_U \cong U \times G\), so \(X|_U \cong U \times Y\) via \([(s,g),y] \mapsto (s, g y)\). Let \((s^i)\) be holomorphic coordinates on \(U\) and \((z^\alpha)\) be local holomorphic coordinates on \(Y\). The connection form \(A\), pulled back to the identity section \((s,e)\), can be written as \[\label{eq:princ95conn95local} A = A_i(s) \mathrm{d}s^i + B_{\bar{j}}(s) \mathrm{d}\bar{s}^j,\tag{32}\] where \(A_i, B_{\bar{j}}: U \to \mathfrak{g}\) are smooth functions. Let \(s^i=x^i+\mathrm{i}y^i\). At \((s,e) \in P\), under the natural identification \(T_{(s,e)} P \cong T_s U \oplus \mathfrak{g}\), the horizontal lift \(\partial_{x^i}^\sharp\) of \(\partial_{x^i}\) is \(\partial_{x^i}^\sharp = \partial_{x^i}-A_i-B_{\bar{i}}\). Similarly, \(\partial_{y^i}^\sharp = \partial_{y^i}-\mathrm{i}A_i+\mathrm{i}B_{\bar{i}}\). Then \[\begin{align} \mathrm{d}q_{(s,e,z)}(\partial_{x^i}^\sharp, 0) &= (\partial_{x^i}, \tilde{\tau}_0(A_i)(z)+\tilde{\tau}_0(B_{\bar{i}})(z)),\\ \mathrm{d}q_{(s,e,z)}(\partial_{y^i}^\sharp, 0) &= (\partial_{y^i}, \tilde{\tau}_0(\mathrm{i}A_i)(z)-\tilde{\tau}_0(\mathrm{i}B_{\bar{i}})(z)). \end{align}\] Therefore, we have \[\begin{align} \nabla_{A}^{1,0}(\partial_i)&=\partial_i+\tfrac{1}{2}\bigl({\tau}_0(A_i)+{\tau}_0(B_{\bar{i}})-\mathrm{i}({\tau}_0(\mathrm{i}A_i)-{\tau}_0(\mathrm{i}B_{\bar{i}}))\bigr) \\&\qquad\,+\tfrac{1}{2}\bigl( \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{{\tau}_0(A_i)} \endgroup +\overline{{\tau}_0(B_{\bar{i}})}-\mathrm{i}( \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{{\tau}_0({\mathrm{i}} A_i)} \endgroup -\overline{{\tau}_0(\mathrm{i}B_{\bar{i}})})\bigr)\\ &=\partial_i+\tau_0(A_i)+\overline{\tau_0(B_{\bar{i}})}\\ \nabla_{A}^{0,1}(\partial_{\bar{i}})&=\partial_{\bar{i}}+\tau_0(B_{\bar{i}})+ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\tau_0(A_i)} \endgroup . \end{align}\] Since \(\im(\tau_0)\subset H^0(Y,TY)\), \(\nabla_A^{1,0}\) is relatively holomorphic and \(\nabla_A^{0,1}\) is mixed relatively holomorphic. ◻

Proof. The map \(A \mapsto \nabla_A^{1,0}\) is affine, since \(\nabla_{A + \alpha}^{1,0}-\nabla_A^{1,0} = \tau(\alpha^{1,0})+ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\tau(\alpha^{0,1})} \endgroup\), which corresponds to \(\tau(\alpha)\in A^1(S,f_*T_{X/S})\). By Lemma 12, the image is contained in \(\mathcal{A}^{1,0}_{\mathrm{RH,LC}}\). It is injective since \(\tau\) is injective. If \(\tau\) is an isomorphism, the map between the modeling spaces \(A^1(S,\ad P)\to A^1(S,f_*T_{X/S})\) is bijective, hence the affine map is an isomorphism. ◻

Proof. Recall that \(F_{D_A}^{1,1}\) is locally given by 21 . Using the notations in the proof of Lemma 12, we compute that \[\begin{align} \bigl( \partial_i \Gamma_{\bar{j}}^\alpha - \partial_{\bar{j}} \Gamma_i^\alpha \bigr) + \bigl( \Gamma_i^\beta \partial_\beta \Gamma_{\bar{j}}^\alpha - \Gamma_{\bar{j}}^\beta \partial_\beta \Gamma_i^\alpha \bigr) &= \tau_0(\partial_i B_{\bar{j}} - \partial_{\bar{j}} A_i)^\alpha + [\tau_0(A_i), \tau_0(B_{\bar{j}})]^\alpha \\ &= \tau_0(\partial_i B_{\bar{j}} - \partial_{\bar{j}} A_i + [A_i, B_{\bar{j}}])^\alpha. \end{align}\] The curvature of \(A\) is \(F_A = \mathrm{d}A + \frac{1}{2}[A, A]\). Its \((1,1)\)-part is locally \[F_A^{1,1} = (\partial_i B_{\bar{j}} - \partial_{\bar{j}} A_i + [A_i, B_{\bar{j}}]) \mathrm{d}s^i \wedge \mathrm{d}\bar{s}^j.\] Thus, \(F^{1,1}_{D_A} = \tau(F_A^{1,1})\). Similarly, \(F^{0,2}_{\bar{\partial}_A} = \tau( F_A^{0,2})\) and \(F^{2,0}_{\partial_A} = \tau(F_A^{2,0})\). Since the lifting condition holds, \(\bar{\partial}_A\) is integrable if and only if \(F^{0,2}_{\bar{\partial}_A}=0\). Since \(\tau\) is injective, this is equivalent to \(F_A^{0,2}=0\). ◻

Proof. Since \(H^0(Y,TY)\) is a finite-dimensional holomorphic representation of the complex reductive group \(G\), the representation is completely reducible. \(\tau_0: \mathfrak{g} \hookrightarrow H^0(Y, TY)\) is a \(G\)-equivariant injection, so \(\mathfrak{g}\) is a sub-representation of \(H^0(Y,TY)\). By complete reducibility, there exists a \(G\)-invariant subspace \(W\) such that \(H^0(Y, TY) \cong \mathfrak{g} \oplus W\). This induces \[f^{\mathrm{hol}}_*T_{X/S} \cong P \times_G (\mathfrak{g} \oplus W) \cong \ad P \oplus (P \times_G W).\] Therefore, \(\tau\) splits. ◻

Proof. Let \(A\) be a principal complex connection on \(P\), then by [2], \(A(P)=[F_A^{1,1}]\) under the Dolbeault isomorphism. By Lemma 14, we have \(\tau([F_A^{1,1}])=[F_{D_A}^{1,1}]_S\). By Lemma 12, \(\nabla^{1,0}_A\) is a relatively holomorphic pure \((1,0)\)-connection. Then \([F_{D_A}^{1,1}]_S=[F_{D}^{1,1}]_S\) by Remark 13. The conclusion then follows. ◻

Proof. Let \(\nabla=\nabla^{1,0}+\nabla^{0,1}\) be a pure complex connection. Then \(\nabla^{1,0}:f^* TS \to TX\) is a pure \((1,0)\)-connection, given locally on \(U_a\) by \[\nabla^{1,0}(\partial_{s_a^i}) = \partial_{s_a^i} + {\Gamma_a}_i^\alpha \partial_{z_a^\alpha}.\] It suffices to show that the Čech \(1\)-cocycle \(c_{ab}\) defined by 34 and the Dolbeault representative \(\mathcal{R}\) locally given by \[\bigl( (\partial_{\bar{s}_a^j} {\Gamma_a}_i^\alpha) \mathrm{d}\bar{s}_a^j+(\partial_{\bar{z}_a^\beta}{\Gamma_a}_i^\alpha)\mathrm{d}\bar{z}_a^\beta \bigr) \otimes \mathrm{d}s_a^i \otimes \partial_{z_a^\alpha}\in A^{0,1}(U_a,f^*T^*S\otimes T_{X/S})\] represent the same cohomology class in \(H^1(X, f^* \Omega_S \otimes T_{X/S})\).

On each \(U_a\), we have the standard holomorphic splitting \(\sigma_a\). Define a Čech \(0\)-cochain \(\gamma = \{\gamma_a\}\) by \[\gamma_a = \nabla^{1,0} - \sigma_a \in C^{\infty}(U_a, f^* T^*S \otimes T_{X/S}).\] Locally on \(U_a\), this is \(\gamma_a = {\Gamma_a}_i^\alpha \mathrm{d}s_a^i \otimes \partial_{z_a^\alpha}\).

Now we compute the Dolbeault differential of \(\gamma_a\). We have \[\bar{\partial}_X \gamma_a = \bar{\partial}_X \bigl( {\Gamma_a}_i^\alpha \mathrm{d}s_a^i \otimes \partial_{z_a^\alpha} \bigr) = \bigl( (\partial_{\bar{s}_a^j} {\Gamma_a}_i^\alpha) \mathrm{d}\bar{s}_a^j+(\partial_{\bar{z}_a^\beta}{\Gamma_a}_i^\alpha)\mathrm{d}\bar{z}_a^\beta \bigr) \otimes \mathrm{d}s_a^i \otimes \partial_{z_a^\alpha}.\] This is exactly the local expression for \(\mathcal{R}|_{U_a}\).

Next, compute the Čech coboundary \(\delta \gamma\). On overlaps \(U_{ab} = U_a \cap U_b\), \[(\delta \gamma)_{ab} = \gamma_b - \gamma_a = (\nabla^{1,0} - \sigma_b) - (\nabla^{1,0} - \sigma_a) = \sigma_a - \sigma_b = c_{ab}.\] Thus \(\delta \gamma = c\), where \(c = \{c_{ab}\}\) is the Čech \(1\)-cocycle.

In the Čech-Dolbeault double complex for the cover \(\mathcal{U}=\{U_a\}\) and the sheaf \(f^* \Omega_S \otimes T_{X/S}\), the element \(\gamma \in \check{C}^0(\mathcal{U}, \mathcal{A}^{0,0}(f^* T^*S \otimes T_{X/S}))\) satisfies \[\mathrm{d}_{\mathrm{tot}} \gamma = \delta \gamma -\bar{\partial}_X \gamma = c - \mathcal{R},\] where \(\mathrm{d}_{\mathrm{tot}}\) is the total differential. Therefore, \(c\) and \(\mathcal{R}\) represent the same cohomology class in \(H^1(X, f^* \Omega_S \otimes T_{X/S})\). ◻

Proof. Fix a point \(s_0 \in S\) and a tangent vector \(v = \frac{\partial}{\partial s_a^i} \big|_{s_0} \in T_{s_0}S\). The extension class \(A(X)\) is represented by the cocycle \(c_{ab}\). Restricting to the fiber \(X_{s_0}\), we obtain a Čech 1-cocycle \(c_{ab}|_{X_{s_0}}(v)\) with values in \(TX_{s_0}\). This represents the extension class of the restricted exact sequence \(0 \to TX_{s_0} \to TX|_{X_{s_0}} \to f^* TS|_{X_{s_0}} \to 0\), contracted by \(v\), which coincides with \(\rho_{s_0}(v)\). Since \(c_{ab}\) depends holomorphically on \(s\), the section \(s \mapsto [c_{ab}|_{X_s}]\) is the global Kodaira-Spencer class \(\mathrm{ks} \in H^0(S, \Omega_S \otimes R^1 f^{\mathrm{hol}}_*T_{X/S})\). Therefore \(\rho(A(X)) = \mathrm{ks}\). ◻

Proof. If \(f\) admits a holomorphic connection, then \(A(X)=0\). By the above remark, \([F_D^{1,1}]_S=0\). By Lemma 15, \(\tau\) induces an injection \(H^{1,1}(S,\ad P)\to H^{1,1}(S,f^{\mathrm{hol}}_*T_{X/S})\). Then by Theorem 18, we have \(A(P)=[F_A^{1,1}]=0\). Therefore, \(\pi_P\) admits a holomorphic connection. ◻

Proof. If such a trivialization exists, then \(\Gamma_i^\alpha = 0\). By 35 , \(R_{ij}^\beta = 0\), so the connection is flat.

Conversely, suppose the connection is flat. The holomorphic horizontal distribution \(H\) is integrable. By the complex Frobenius theorem, \(H\) defines a holomorphic foliation \(\mathcal{F}\) transverse to the fibers. Locally around any point \(x \in X\), we can find adapted holomorphic coordinates \((s, z)\) such that the leaves of the foliation are given by \(\{z = \text{constant}\}\) and the projection \(f\) is locally given by \((s, z) \mapsto s\). In these coordinates, the horizontal distribution \(H\) is spanned by \(\partial_i\), which means \(\Gamma_i^\alpha\) must be identically zero. ◻

Proof. It is well-known that principal connections are complete (see [23]). Let \(q: P \times Y \to X\) be the quotient map. Recall that the horizontal distribution \(H_X^\RR\) of \(\nabla_A^\RR\) is the image under \(\mathrm{d}q\) of the distribution \(H_P^\RR \oplus 0\) on \(P \times Y\), where \(H_P^\RR\) is the horizontal distribution of \(A\).

Let \(x_0 \in X_{\gamma(0)}\). We can represent \(x_0\) as \([p_0, y_0]\) for some \(p_0 \in P_{\gamma(0)}\) and \(y_0 \in Y\). The horizontal lift \(\tilde{\gamma}_P(t)\) of \(\gamma(t)\) in \(P\) starting at \(p_0\) exists for all \(t \in [0, 1]\). Consider the path \(\tilde{\gamma}_X(t)\) in \(X\) defined by \[\tilde{\gamma}_X(t) := q(\tilde{\gamma}_P(t), y_0) = [\tilde{\gamma}_P(t), y_0].\] We verify that this is the required horizontal lift. Clearly, \(f(\tilde{\gamma}_X(t)) = \pi_P(\tilde{\gamma}_P(t)) = \gamma(t)\), and \(\tilde{\gamma}_X(0) = [p_0, y_0] = x_0\). The tangent vector is \(\tilde{\gamma}_X'(t) = \mathrm{d}q_{(\tilde{\gamma}_P(t), y_0)} (\tilde{\gamma}_P'(t), 0)\). Since \(\tilde{\gamma}_P(t)\) is horizontal in \(P\), \(\tilde{\gamma}_P'(t) \in H_P^\RR\). Thus, \((\tilde{\gamma}_P'(t), 0) \in H_P^\RR \oplus 0\). By the definition of \(H_X^\RR\), \(\tilde{\gamma}_X'(t)\) lies in \(H_X^\RR\). Since the horizontal lift \(\tilde{\gamma}_X(t)\) is defined for all \(t \in [0, 1]\), the induced connection \(\nabla_A^{1,0}\) is complete. ◻

Proof. It suffices to work in a local holomorphic chart \((s, z)\), where the connection is given by smooth coefficients \(\Gamma_i^\alpha(s, z)\) and \(\Gamma_i^{\bar{\alpha}}(s,z)\). A path \((s(t), z(t))\) in \(X\) is horizontal if 37 is satisfied. Let \(z(t; z_0)\) be the solution to 37 with initial condition \(z_0\). Let \(w^\alpha_\beta(t) = \frac{\partial z^\alpha}{\partial \bar{z}_0^\beta}(t)\). Then \(w^\alpha_\beta(0)=0\). Differentiate 37 with respect to \(\bar{z}_0^\beta\), \[\label{eq:dbarz95ODE} \frac{\mathrm{d}w^\alpha_\beta}{\mathrm{d}t} = \biggl( \frac{\partial \Gamma_i^\alpha}{\partial z^\gamma} w^\gamma_\beta + \frac{\partial \Gamma_i^\alpha}{\partial \bar{z}^\gamma} \frac{\partial \bar{z}^\gamma}{\partial \bar{z}_0^\beta} \biggr) \frac{\mathrm{d}s^i}{\mathrm{d}t}+\biggl( \frac{\partial \overline{\Gamma_i^{\bar\alpha}}}{\partial z^\gamma} w^\gamma_\beta + \frac{\partial \overline{\Gamma_i^{\bar\alpha}}}{\partial \bar{z}^\gamma} \frac{\partial \bar{z}^\gamma}{\partial \bar{z}_0^\beta} \biggr) \frac{\mathrm{d}\bar{s}^i}{\mathrm{d}t}.\tag{38}\]

If \(\nabla^{1,0}\) is relatively holomorphic and \(\begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\nabla^{1,0}} \endgroup\) is mixed relatively holomorphic, then 38 simplifies to \[\frac{\mathrm{d}w^\alpha_\beta}{\mathrm{d}t} = \biggl(\frac{\partial \Gamma_i^\alpha}{\partial z^\gamma} \frac{\mathrm{d}s^i}{\mathrm{d}t}+\frac{\partial \overline{\Gamma_i^{\bar\alpha}}}{\partial z^\gamma} \frac{\mathrm{d}\bar{s}^i}{\mathrm{d}t}\biggr) w^\gamma_\beta .\] By uniqueness of solutions, \(w^\alpha_\beta(t) = 0\) for all \(t\). Thus, \(\tau_t\) is holomorphic. Since \(\tau_t\) is bijective, it is a biholomorphism.

Conversely, suppose that for any smooth path \(\gamma(t)\) in \(S\), the parallel transport map \(\tau_t: z_0 \mapsto z(t)\) is holomorphic. This implies that \(w^\alpha_\beta(t) = 0\), and then \(\frac{\mathrm{d}w^\alpha_\beta}{\mathrm{d}t}(0) = 0\). By 38 , \[0 = \frac{\partial \Gamma_i^\alpha}{\partial \bar{z}^\beta}(s(0), z_0) \frac{\mathrm{d}s^i}{\mathrm{d}t}(0)+\frac{\partial \overline{\Gamma_i^{\bar \alpha}}}{\partial \bar{z}^\beta}(s(0), z_0) \frac{\mathrm{d}\bar{s}^i}{\mathrm{d}t}(0).\] Since we can choose a path passing through any point \((s(0), z_0)\) with any velocity, we must have \[\frac{\partial \Gamma_i^\alpha}{\partial \bar{z}^\beta} = \frac{\partial \overline{\Gamma_i^{\bar \alpha}}}{\partial \bar{z}^\beta} = 0. \qedhere\] ◻

Proof. Let \(U\) be a holomorphic coordinate ball centered at \(s_0\). Define \(\Phi: U \times Y \to f^{-1}(U)\) by \(\Phi(s, z_0) = \tau(\gamma_s)(z_0)\), where \(\gamma_s(t)=ts\) is the radial path. The ODE 36 becomes \[\label{eq:ODE95with95parameters} \frac{\mathrm{d}z^\alpha}{\mathrm{d}t} = \Gamma_i^\alpha(ts, z(t)) s^i.\tag{39}\] Since \(\Gamma_i^\alpha\) are holomorphic functions, by holomorphic dependence on parameters, the solution \(z(t; s, z_0)\) is holomorphic in \((s, z_0)\). Thus \(\Phi\) is a holomorphic map. By Lemma 20, its restriction to \(\{s\} \times Y\) is a biholomorphism onto \(X_s\). So \(\Phi\) is a bijection, and is a biholomorphism as desired. ◻

Proof. Let \(\mathcal{F}\) be the holomorphic transversal foliation determined by \(\nabla^{1,0}\). Let \(H(t, u): [0, 1] \times [0, 1] \to S\) be a smooth homotopy between \(\gamma_0\) and \(\gamma_1\). Fix \(x_0 \in X_{s_0}\). Let \(\widetilde{H}(t, u)\) be the unique horizontal lift of \(\gamma_u(t):=H(t,u)\) starting at \(x_0\). By the smooth dependence of solutions of ODEs on parameters, the map \(\widetilde{H}: [0, 1] \times [0, 1] \to X\) is smooth.

Let \(L_{x_0}\) be the unique leaf of the foliation \(\mathcal{F}\) passing through \(x_0\). By construction, \[\widetilde{H}([0, 1] \times [0, 1]) \subset L_{x_0}.\] Define the path \(\beta: [0, 1] \to X\) by \(\beta(u) = \widetilde{H}(1, u)\). We have \[f(\beta(u)) = f(\widetilde{H}(1, u)) = H(1, u) = s_1.\] Thus, the path \(\beta(u)\) lies in the intersection \(L_{x_0} \cap X_{s_1}\), which is discrete since the foliation \(\mathcal{F}\) is transverse to the fibers. The map \(\beta(u)\) is continuous, hence must be constant. In particular, \(\beta(0) = \beta(1)\). Since \(x_0\) is arbitrary, \(\tau(\gamma_0) = \tau(\gamma_1)\). ◻

Proof. Let \(\{U_a\}\) be a cover of \(S\) by coordinate balls. Let \(s_a \in U_a\) be the center and identify \(Y = X_{s_a}\). Since \(\nabla^{1,0}\) is complete and holomorphic, we obtain a local holomorphic trivialization \(\Phi_a: U_a \times Y \to f^{-1}(U_a)\) by Lemma 21. This collection of charts forms a holomorphic atlas for \(X\).

If we fix \(z_0 \in Y\), the map \(s \mapsto \Phi_a(s, z_0)\) defines a local section. Due to path independence, this section traces out the unique leaf of the foliation \(\mathcal{F}\) passing through \(z_0\) at \(s_a\). By definition, this is a horizontal section. Therefore, in the coordinates induced by \(\Phi_a\), the horizontal sections are precisely the constant sections \(\{z_a = \text{constant}\}\). A trivialization with this property is called a flat trivialization.

Let \(U_{ab} = U_a \cap U_b\), \(g_{ab}: U_{ab}\to \Aut(Y)\) be the transition map, given by \[g_{ab}(s) = \Phi_a(s, \cdot)^{-1} \comp \Phi_b(s, \cdot).\] Fix \(z_b \in Y\). The map \(s \mapsto (s, z_b)\) represents a horizontal section in the \(\Phi_b\)-trivialization. In the \(\Phi_a\)-trivialization, this horizontal section is represented by \((s,g_{ab}(s)(z_b))\). Hence \(g_{ab}(s)\) is locally constant. ◻

Proof. Consider a representation \(\rho: \pi_1(S,s_0) \to G=\Aut(Y)\). Let \(\pi: \widetilde{S} \to S\) be the universal covering of \(S\). Define an action of \(\pi_1(S,s_0)\) on the product \(\widetilde{S} \times Y\) by \[\gamma\cdot (\tilde{s}, y) = (\gamma\cdot \tilde{s}, \rho(\gamma)y), \quad \gamma \in \pi_1(S,s_0).\] Since the actions on \(\widetilde{S}\) and \(Y\) are holomorphic, this diagonal action is holomorphic, free, and properly discontinuous. The quotient space \(X := (\widetilde{S} \times Y)/\pi_1(S,s_0)\) (denoted \(\widetilde{S} \times_\rho Y\)) is a complex manifold. The projection \(f: X \to S\) induced by \(\pi\) is a holomorphic fiber bundle with fiber \(Y\).

We define a flat holomorphic connection on \(X\). Consider the trivial connection \(\nabla_0\) on the projection \(\widetilde{S} \times Y \to \widetilde{S}\). Its horizontal distribution is \(H_0 = T\widetilde{S} \oplus 0 \subset T(\widetilde{S} \times Y)\), which is a holomorphic subbundle. Let \(L_\gamma\) denote the action of \(\gamma\). The differential acts on a horizontal vector \((v, 0) \in H_0(\tilde{s}, y)\) as \[\mathrm{d}L_\gamma (v, 0) = (\mathrm{d}( \gamma \cdot ) v, \mathrm{d}(\rho(\gamma)) 0) = (\mathrm{d}(\gamma\cdot) v, 0).\] This vector is again horizontal. Thus, \(H_0\) descends to a holomorphic distribution \(H\) on \(X\), defining a holomorphic connection \(\nabla^{1,0}\) on \(f: X \to S\). Since the curvature of the trivial connection \(\nabla_0\) is zero, and the curvature is a tensor that descends to the quotient, the curvature of \(\nabla^{1,0}\) is also zero. Moreover, \(\nabla^{1,0}\) is complete. Let \(\gamma: [0, 1] \to S\) be a path and \(x_0 \in X_{\gamma(0)}\). Lift \(\gamma\) to a path \(\tilde{\gamma}\) in the universal cover \(\widetilde{S}\). We can represent \(x_0\) as \([(\tilde{\gamma}(0), y_0)]\) for some \(y_0 \in Y\). The horizontal lift of \(\gamma\) starting at \(x_0\) is the projection of the trivial horizontal lift in \(\widetilde{S} \times Y\), given explicitly by \(t \mapsto [(\tilde{\gamma}(t), y_0)]\). Since \(\tilde{\gamma}(t)\) is defined for all \(t \in [0, 1]\), the horizontal lift exists for the entire interval, proving that \(\nabla^{1,0}\) is complete.

Conversely, let \((X, \nabla^{1,0})\) be a complete flat holomorphic fiber bundle over \(S\) with fiber \(Y\). Fix a base point \(s_0 \in S\) and identify the fiber \(X_{s_0}\) with \(Y\). For any loop \(\gamma\) based at \(s_0\), the parallel transport \(\tau(\gamma): Y \to Y\) is defined by lifting \(\gamma\) horizontally. By Lemma 20, \(\tau(\gamma)\) is a biholomorphism, i.e., \(\tau(\gamma) \in G\). Since \(\nabla^{1,0}\) is flat, \(\tau(\gamma)\) depends only on the homotopy class of \(\gamma\). This defines the monodromy representation \[\rho: \pi_1(S, s_0) \to G, \quad \rho([\gamma]) = \tau(\gamma)^{-1}.\]

We verify that these constructions are inverse to each other up to isomorphism. Suppose we start with a representation \(\rho\) and construct the complete flat holomorphic bundle \((X, \nabla^{1,0}) = \widetilde{S} \times_\rho Y\). Fix \(s_0 \in S\) and a lift \(\tilde{s}_0 \in \widetilde{S}\). Identify \(Y \cong X_{s_0}\) via \(y \mapsto [(\tilde{s}_0, y)]\). A loop \(\gamma\) lifts to a path from \(\tilde{s}_0\) to \([\gamma]\cdot\tilde{s}_0\). The parallel transport of \(y\) along \(\gamma\) is \(t \mapsto [(\tilde{\gamma}(t), y)]\). The endpoint is \([([\gamma]\cdot\tilde{s}_0, y)] = [(\tilde{s}_0, \rho([\gamma])^{-1}y)]\). Thus the holonomy of \((X, \nabla^{1,0})\) recovers \(\rho\).

Suppose we start with a complete flat holomorphic bundle \((X, \nabla^{1,0})\), and let \(\rho\) be its monodromy representation. We construct an isomorphism \(\Psi: \widetilde{S} \times_\rho Y \to X\). Let \(\alpha\) be a path in \(\widetilde{S}\) from \(\tilde{s}_0\) to \(\tilde{s}\), and \(p(\alpha)\) its projection in \(S\). Define \(\Phi: \widetilde{S} \times Y \to X\) by \(\Phi(\tilde{s}, y) = \tau(p(\alpha))(y)\), using the identification \(Y\cong X_{s_0}\). \(\Phi\) is holomorphic because \(\nabla^{1,0}\) is holomorphic and flat (so \(\Phi\) is well-defined). \(\Phi\) maps the trivial horizontal distribution on \(\widetilde{S} \times Y\) to the horizontal distribution of \(\nabla^{1,0}\). One verifies that \(\Phi\) respects the action of \(\pi_1(S,s_0)\) using the definition of \(\rho\), so it descends to an isomorphism \(\Psi\).

It remains to show that for any two complete flat bundles \((X_1, \nabla_1^{1,0})\) and \((X_2, \nabla_2^{1,0})\) with fibers \(Y_1, Y_2\) over \(s_0\) and corresponding monodromy representations \(\rho_1, \rho_2\), the map \[\mathsf{F}_{1,2}: \Hom_{\mathbf{CFB}}((X_1, \nabla_1^{1,0}), (X_2, \nabla_2^{1,0})) \to \Hom_{\mathbf{REP}}((Y_1, \rho_1), (Y_2, \rho_2))\] given by restricting a bundle morphism to the fiber at \(s_0\) (\(F \mapsto F|_{X_{1, s_0}}\)) is a bijection.

Injectivity: Suppose \(F, G: X_1 \to X_2\) are two morphisms such that \(F|_{X_{1, s_0}} = G|_{X_{1, s_0}}\). Let \(x \in X_1\) be any point. Since \(S\) is connected and the connection is complete, there exists a horizontal path \(\tilde{\gamma}\) in \(X_1\) connecting some \(y \in X_{1, s_0}\) to \(x\). Let \(\gamma\) be the projection of this path on \(S\). Since morphisms preserve horizontal distributions, \(F(\tilde{\gamma})\) and \(G(\tilde{\gamma})\) must be the unique horizontal lifts of \(\gamma\) in \(X_2\) starting at \(F(y)\) and \(G(y)\) respectively. Since \(F(y) = G(y)\), uniqueness implies \(F(x) = G(x)\). Thus \(F=G\).

Surjectivity: Let \(\phi: Y_1 \to Y_2\) be a morphism in \(\mathbf{REP}(\pi_1(S,s_0))\). We construct a bundle morphism \(F\). Recall that \(X_i \cong (\widetilde{S} \times Y_i)/\pi_1(S,s_0)\). Define \[\widetilde{F}: \widetilde{S} \times Y_1 \to \widetilde{S} \times Y_2, \quad (\tilde{s}, y) \mapsto (\tilde{s}, \phi(y)).\] \(\widetilde{F}\) is holomorphic and preserves the trivial horizontal distribution (as it acts as identity on \(\widetilde{S}\)). We verify it descends to the quotients. For any \(\gamma \in \pi_1(S,s_0)\), \[\widetilde{F}(\gamma \cdot (\tilde{s}, y)) = \widetilde{F}(\gamma\tilde{s}, \rho_1(\gamma)y) = (\gamma\tilde{s}, \phi(\rho_1(\gamma)y))=\gamma \cdot (\tilde{s}, \phi(y)) = \gamma \cdot \widetilde{F}(\tilde{s}, y),\] by the \(\pi_1(S,s_0)\)-equivariance of \(\phi\). Thus \(\widetilde{F}\) descends to a holomorphic bundle morphism \(F: X_1 \to X_2\). Since \(\widetilde{F}\) preserves the trivial horizontal distributions, \(F\) preserves the induced horizontal distributions. By construction, \(F|_{X_{1, s_0}} = \phi\). ◻

Proof. By Theorem 20, \(f: X \to S\) admits a holomorphic connection, which is complete as \(Y\) is compact. Let \(\nabla_{\mathrm{hol}}\) be such a connection. Its curvature \(F_{\nabla_{\mathrm{hol}}}\) is of type \((2,0)\) which must vanish since \(S\) is a Riemann surface. Then the statement follows from Theorem 26. ◻

Proof. The flat \(G\)-bundle \(P\) over \(S\) is determined (up to isomorphism) by a homomorphism \(\rho: \pi_1(S) \to G\). Let \(g\) be the genus of \(S\). The fundamental group of \(S\) is generated by \(a_1, b_1, \dots, a_g, b_g\) subject to the relation \(\prod_{i=1}^g [a_i, b_i] = e\), where \([x, y] = xyx^{-1}y^{-1}\) denotes the commutator. Since \(\rho\) is a homomorphism, the images \(A_i = \rho(a_i)\) and \(B_i = \rho(b_i)\) in \(G\) satisfy the relation \(\prod_{i=1}^g [A_i, B_i] = e_G\). Let \(\widetilde{G}\) be the universal cover of \(G\). By [24], \[c(P) = \prod_{i=1}^g [\widetilde{A}_i, \widetilde{B}_i],\] where \(\widetilde{A}_i, \widetilde{B}_i \in \widetilde{G}\) are arbitrary lifts of \(A_i, B_i\). The characteristic element is given by the product of their commutators in \(\widetilde{G}\), which is well-defined since the kernel \(\pi_1(G)\) of \(\widetilde{G}\to G\) is central.

Since \(G\) is a connected complex reductive group, \(\widetilde{G}=\widetilde{Z}_0\times \widetilde{G}'\), where \(\widetilde{G}'\) is the universal cover of the semisimple part \(G'=[G,G]\), \(\widetilde{Z}_0\) is the universal cover of the identity component \(Z_0\) of the center of \(G\). We have \(c(P) \in \{e\} \times \widetilde{G}'\). Since \(\pi_1(G')\) is finite [25], \(c(P)\) is a torsion element in \(\pi_1(G)\). ◻

Proof. By Corollary 6, \(f\) admits a holomorphic connection if \(\pi_P\) does. Let \(\chi: G \to \mathbb{C}^*\) be a holomorphic character. It induces a homomorphism \(\chi_*: \pi_1(G) \to \pi_1(\mathbb{C}^*) \cong \mathbb{Z}\). By [24], \(\deg(L_\chi) = \chi_*(c(P))\), where \(L_\chi = P \times_\chi \mathbb{C}^*\). Assume \(c(P)\) is a torsion element in \(\pi_1(G)\). There exists a positive integer \(k\) such that \(k \cdot c(P) = 0\). For any \(\chi\), we have \[k \cdot \deg(L_\chi) = k \cdot \chi_*(c(P)) = \chi_*(k \cdot c(P)) = \chi_*(0) = 0.\] Then \(\deg(L_\chi) = 0\). Note that by Weil’s theorem, \(\ad P\) admits a holomorphic connection if and only if each summand in the Remak decomposition of \(\ad P\) has degree zero. If moreover \(\ad P\) admits a holomorphic connection, then so does \(\pi_P\) by [15].

Suppose \(H^0(Y,TY)\) is finite-dimensional, and \(f\) admits a holomorphic connection. By Corollary 7, \(P\) also admits a holomorphic connection which is automatically flat on a Riemann surface. By Lemma 23, \(c(P)\) is torsion. By [15], \(\ad P\) also admits a holomorphic connection. ◻

Proof. This follows from Lemma 9 and the fact that if \(u\) and \(Ju\) are infinitesimal automorphisms of a Kähler manifold, then \(u\) is parallel [27]. ◻

Proof. By Lemma 27, it suffices to work locally on a unitary trivialization. By assumption we may write \[\bar{\partial}_f-\bar{\partial}_a=v+\mathrm{i}w,\] where \(v,w\in A^{0,1}(U_a,\mathfrak{aut}(Y,\omega_Y))\) (possibly in a non-unique way). We define an almost connection \(\partial\) by \[\label{eq:kah95al95conn95loc} (\partial-\partial_a)(\partial_i)=v(\partial_{\bar{i}})-\mathrm{i}w(\partial_{\bar{i}}),\tag{44}\] where \(\partial_a\) is the trivial almost connection determined by the unitary trivialization. Then \(\hat{D}=\partial+\bar{\partial}_f\) corresponds to a Kähler connection on \(U_a\). Using a partition of unity subordinate to \(\{U_a\}\), we obtain a global Kähler connection \(\nabla^{1,0}\).

Conversely, let \(\nabla^{1,0}\) be a complete Kähler connection, which is equivalent to \(\partial+\bar{\partial}_f\). Using the radial parallel transports as in Lemma 21, we obtain a unitary atlas. On each unitary atlas \(U_a\), we have \((\partial-\partial_a)(\partial_i)+(\bar{\partial}_f-\bar{\partial}_a)(\partial_{\bar{i}})\in \mathfrak{aut}(Y,\omega_Y)\) and \((\partial-\partial_a)(\partial_i)-(\bar{\partial}_f-\bar{\partial}_a)(\partial_{\bar{i}})\in\mathrm{i}\, \mathfrak{aut}(Y,\omega_Y)\). Thus, \(\bar{\partial}_f-\bar{\partial}_a\in A^{0,1}(U_a,\mathfrak{aut}(Y,\omega_Y)_\CC)\). ◻

Proof. It suffices to prove that the local construction as in the proof of the above proposition is independent of the choice of such unitary trivializations. Suppose there are two unitary trivializations \((U_a,\Phi_a)\) and \((U_b,\Phi_b)\), such that \(\bar{\partial}_f-\bar{\partial}_a \in A^{0,1}(U_a,\mathfrak{k}^\CC)\) and \(\bar{\partial}_f-\bar{\partial}_b \in A^{0,1}(U_b,\mathfrak{k}^\CC)\). We have \(p+\mathrm{i}q:=\bar{\partial}_a-\bar{\partial}_b\in A^{0,1}(U_{ab},\mathfrak{k}^\CC)\), where \(p,q\in A^{0,1}(U_{ab},\mathfrak{k})\). Then \((\partial_a-\partial_b)(\partial_i)=p(\partial_{\bar{i}})-\mathrm{i}q(\partial_{\bar{i}})\). Let \(\partial_{U_a}\) and \(\partial_{U_b}\) be the local almost connections constructed as in 44 , then \[\begin{align} (\partial_{U_b}-\partial_{U_a})(\partial_i)&=\bigl((\partial_{U_b}-\partial_b)-(\partial_{U_a}-\partial_a)-(\partial_a-\partial_b)\bigr)(\partial_i)\\ &=\bigl((v+p)-\mathrm{i}(w+q)-(v-\mathrm{i}w)-(p-\mathrm{i}q)\bigr)(\partial_i)=0, \end{align}\] where \(\bar{\partial}_f-\bar{\partial}_a=:v+\mathrm{i}w\), \(v,w\in A^{0,1}(U_a,\mathfrak{k})\). ◻

Proof. We work in a local unitary trivialization as in the above proposition, where \(\bar{\partial}_f\) can be expressed as \(\partial_{\bar{i}}\mapsto \partial_{\bar{i}}+(v_i+\mathrm{i}w_i)\mod \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup\), where \(v_i,w_i\in \mathfrak{aut}(Y,\omega_Y)\). Then \(\partial\) is given by \(\partial_i\mapsto \partial_i+(v_i-\mathrm{i}w_i)\mod \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{T_{X/S}} \endgroup\). We compute \[\begin{align} F_{\bar{\partial}_f}^{0,2}(\partial_{\bar{i}},\partial_{\bar{j}})&=[\partial_{\bar{i}}+v_i+\mathrm{i}w_i,\partial_{\bar{j}}+v_j+\mathrm{i}w_j]\\ &=\partial_{\bar{i}}(v_j+\mathrm{i}w_j)+[v_i+\mathrm{i}w_i,v_j+\mathrm{i}w_j]-\partial_{\bar{j}}(v_i+\mathrm{i}w_i)\\ &=\tfrac{1}{2}(\partial_{x^i}v_j-\partial_{y^i}w_j)+[v_i,v_j]-[w_i,w_j]-\tfrac{1}{2}(\partial_{x^j}v_i-\partial_{y^j}w_i)\\ &\quad +\mathrm{i}\bigl(\tfrac{1}{2}(\partial_{y^i}v_j+\partial_{x^i}w_j)+[w_i,v_j]+[v_i,w_j]-\tfrac{1}{2}(\partial_{y^j}v_i+\partial_{x^j}w_i)\bigr), \end{align}\] where \(s^i=x^i+\mathrm{i}y^i\). On the other hand, \[\begin{align} F_{\partial}^{2,0}(\partial_{i},\partial_{j})&=[\partial_{i}+v_i-\mathrm{i}w_i,\partial_{j}+v_j-\mathrm{i}w_j]\\ &=\partial_{i}(v_j-\mathrm{i}w_j)+[v_i-\mathrm{i}w_i,v_j-\mathrm{i}w_j]-\partial_j(v_i-\mathrm{i}w_i)\\ &=\tfrac{1}{2}(\partial_{x^i}v_j-\partial_{y^i}w_j)+[v_i,v_j]-[w_i,w_j]-\tfrac{1}{2}(\partial_{x^j}v_i-\partial_{y^j}w_i)\\ &\quad -\mathrm{i}\bigl(\tfrac{1}{2}(\partial_{y^i}v_j+\partial_{x^i}w_j)+[w_i,v_j]+[v_i,w_j]-\tfrac{1}{2}(\partial_{y^j}v_i+\partial_{x^j}w_i)\bigr). \end{align}\] The result follows. ◻

Proof. Since the trivialization \(\Phi_a\) is induced by a local section \(\sigma_a: U_a \to P\), we have \(\Phi_{a, s}^{-1}(y) = [\sigma_a(s), y] = \phi_{\sigma_a(s)}(y)\). The unitary condition means \(\omega_s = \Phi_{a, s}^* \omega_Y\). Then we get \(\phi_{\sigma_a(s)}^* (\omega_s) = \omega_Y\). Now let \(p \in P_s\) be arbitrary, we can write \(p = \sigma_a(s) \cdot g\) for some \(g\in G\). Then \[\phi_p^* (\omega_s) = (\phi_{\sigma_a(s)} \comp g)^* (\omega_s) = g^* \phi_{\sigma_a(s)}^* (\omega_s)= g^*\omega_Y\in G\cdot \omega_Y. \qedhere\] ◻

Proof. \(\Rightarrow\): Let \(\omega_s :=(\phi_p^{-1})^*\omega_Y\), for \(p\in (P_K)_s\). We verify this is well-defined. Let \(p, p' \in (P_K)_s\). There is a unique \(k\in K\) such that \(p' = p\cdot k\). We have \(\phi_{p'} = \phi_p \comp k\). Then \[\phi_{p'}^* \omega_s = (\phi_p \comp k)^* \omega_s = k^* (\phi_p^* \omega_s) = k^* \omega_Y=\omega_Y.\] The metric \(\omega_s\) is well-defined on each fiber and is \(G\)-modeled on \((Y, \omega_Y)\). Since \(P_K\) is a smooth subbundle, this construction yields a smooth fiberwise Kähler metric \(\omega_{X/S}\).

\(\Leftarrow\): Suppose \(\omega_{X/S}\) is a fiberwise Kähler metric \(G\)-modeled on \((Y, \omega_Y)\). Define \[P_H := \{p \in P \mid \phi_p^* \omega_s = \omega_Y, \text{ where } s=\pi_P(p) \}.\] Let \(p \in P_H\) and \(g \in G\), then \(\phi_{p\cdot g}^* \omega_s = g^* (\phi_p^* \omega_s) = g^* \omega_Y\). Thus, \(p\cdot g \in P_H\) if and only if \(g^* \omega_Y = \omega_Y\), which is equivalent to \(g \in H\). Therefore, the right action of \(H\) on \(P_H\) is free and transitive on the fibers, and \(P_H\) is a principal \(H\)-subbundle of \(P\).

As \(K \subset H \subset G\), \(K\) is also a maximal compact subgroup of \(H\). A reduction of \(P_H\) from \(H\) to \(K\) is equivalent to a global section of the associated bundle \(P_H \times_H (H/K) \to S\). Since the fiber \(H/K\) is contractible, such a section always exists. Now, let \(P_K \subset P_H\) be any such \(K\)-reduction which is also a \(K\)-reduction of the original bundle \(P\). Let \(\omega'_{X/S}\) be the fiberwise Kähler metric induced by \(P_K\) according to the first part of the lemma. By the construction, \(\phi_p^* \omega_s = \omega_Y=\phi_p^* \omega'_s\) for any \(p \in (P_K)_s\) and \(s \in S\). Therefore, \(\omega'_{X/S} = \omega_{X/S}\). ◻

Proof. Clearly, the induced connection \(\nabla_A^{1,0}\) is a Kähler connection since the parallel transport maps are Kähler isometries. By the construction of \(\omega_{X/S}\) in Lemma 29, any atlas of \(P_K\) induces a unitary atlas of \((f,\omega_{X/S})\) such that the conditions of Proposition 29 hold. Therefore, the Kähler connection \(\nabla_A^{1,0}\) is the Chern connection. ◻

Proof. This essentially follows from [26] and [29], we just recall the construction of \(\omega\). On \(P_K\times Y\), define the 2-form \[\widehat\omega := \omega_Y - \mathrm{d}\langle\mu,A_K\rangle.\] This form is closed. One can show that \[\label{eq:symplectic95coupling} \widehat\omega = \pi_{A_K}^\ast\omega_Y - \langle \mu, F_{A_K}\rangle,\tag{48}\] where \(\pi_{A_K}: TP_K^{\RR}\times TY^{\RR} \to TY^{\RR}\) is given by \(\pi_{A_K}(v,\hat{y}) = \hat{y} + v_{A_K(v)}(y)\), and \(F_{A_K}\in A^2(P_K,\mathfrak{k})\) is the curvature of \(A_K\). Then \(\widehat\omega\) is basic and descends to a smooth closed \((1,1)\)-form \(\omega\) on \(X=P_K\times_K Y\). For each \(s\in S\), the restriction \(\omega|_{X_s}\) can be identified with \(\omega_Y\), hence \(\omega\) is a relatively Kähler form on \(X\).

Next, we verify that for any \(s\in S\) and \(p \in (P_K)_s\), \(\phi_p^* (\omega|_{X_s}) = \omega_Y\). We have \(\phi_p = q \comp i_p\), where \(i_p: Y \to P_K\times Y\) is the inclusion \(i_p(y) = (p,y)\), and \(q: P_K\times Y \to X\) is the quotient map. By construction, \(q^* \omega = \widehat\omega\), so \(\phi_p^* \omega = i_p^* \widehat\omega=\omega_Y\). ◻

Proof. We verify this locally. In a local holomorphic trivialization \(X|_U \cong U \times Y\) with coordinates \((s^i, z^\alpha)\), the complex connection \(A\) is locally given by a \(\mathfrak{g}\)-valued \((1,0)\)-form \(A = A_i(s) \mathrm{d}s^i\). By the definition of \(\nabla_A^{1,0}\), the corresponding horizontal lift is \[\label{eq:assoc95conn95lift} \nabla_A^{1,0}(\partial_{s^i}) = \partial_{s^i} + v_{A_i(s)},\tag{49}\] where \(v_{A_i(s)}=\tau_0(A_i(s))\) is the holomorphic vector field on \(Y\) generated by \(A_i(s)\).

Now we compute \(\nabla_\omega^{1,0}\). We need the mixed components \(g_{i\bar\beta}\) of \(\omega\) (cf. Eq. 46 ). Since \(\widehat{\omega}=\omega_Y-\mathrm{d}\langle\mu,A_K\rangle\) and \(A\) is induced by \(A_K\), we have \[\label{eq:mixed-coeff95generalized} \mathrm{i}g_{i\bar\beta} = \partial_{\bar\beta} \langle\mu, A_i\rangle,\tag{50}\] where we extended the moment map pairing \(\langle \mu, \cdot \rangle\) complex-linearly to \(\mathfrak{g} = \mathfrak{k}^\CC\). By the Hamiltonian identity, \(\iota_{v_\xi}\omega_Y = -\bar\partial\langle\mu,\xi\rangle\), since \(v_\xi\) is a holomorphic vector field and \(\omega_Y\) is of type \((1,1)\). Locally, this implies \[\partial_{\bar\beta} \langle\mu,\xi\rangle = -\mathrm{i}g_{\alpha\bar\beta} v_\xi^\alpha.\] Combining this with 50 for \(\xi = A_i\), we obtain \[g_{i\bar\beta} = -g_{\alpha\bar\beta} v_{A_i}^\alpha.\] By 45 , \(\Gamma_i^\alpha=-g^{\bar{\beta}\alpha}g_{i\bar{\beta}}=v_{A_i}^\alpha\). Therefore, \(\nabla_\omega^{1,0}(\partial_{s^i})=\partial_{s^i} + v_{A_i}^\alpha \partial_\alpha = \partial_{s^i} + v_{A_i}\), which coincides with 49 . ◻

Proof. This follows immediately from Lemma 31 and Lemma 12. ◻

Proof. By Lemma 14, \(G_{D''}^{2,0}=\tau(G_{(\bar{\partial}_\pi,\theta)}^{2,0})\). Let \(u, v \in TS\), we have \[G_{D''}^{2,0}(u, v) = [\tau(\theta(u)), \tau(\theta(v))] = \tau([\theta(u),\theta(v)])=\tau\!\left(\tfrac12[\theta,\theta](u,v)\right)=\tau(G_{(\bar{\partial}_\pi,\theta)}^{2,0}(u,v)).\] \(G_{D''}^{1,1}\) is locally given by 52 , we compute \[\begin{align} -\partial_{\bar{j}} \tau_0(\theta_i)^\alpha + \tau_0(\theta_i)^\gamma \partial_\gamma \Gamma_{\bar{j}}^\alpha- \Gamma_{\bar{j}}^\gamma \partial_\gamma \tau_0(\theta_i)^\alpha&=\tau_0(-\partial_{\bar{j}}\theta_i)^\alpha-[\tau_0(B_{\bar{j}}), \tau_0(\theta_i)]^\alpha\\ &=-\tau_0(\partial_{\bar{j}}\theta_i+[B_{\bar{j}},\theta_i])^\alpha. \end{align}\] On the other hand, \[G_{(\bar{\partial}_\pi,\theta)}^{1,1}=(\mathrm{d}_A\theta)^{1,1}=\bar{\partial}\theta+[B_{\bar{j}},\theta_i]\mathrm{d}\bar{s}^j\wedge \mathrm{d}s^i=-(\partial_{\bar{j}}\theta_i+[B_{\bar{j}},\theta_i])\mathrm{d}s^i\wedge\mathrm{d}\bar{s}^j.\] Therefore, \(G_{D''}^{1,1}=\tau(G_{(\bar{\partial}_\pi,\theta)}^{1,1})\). ◻

Proof. Since \(\tau\) is injective, we may identify \(\theta\) with \(\tau^{-1}(\theta)=\theta_P\in A^{1,0}(S,\ad P)\). In this case, \(\mathfrak{k}_{X/S}=\ad P_K\) and \(\mathfrak{k}_{X/S}^\CC=\ad P\). Then by Definition 23, \(\bar{\theta}_{\omega_{X/S}}\) is identified with \(\bar{\theta}_{P,K}\) via \(\tau\). ◻

Proof. By the definition of the quotient \(X\), \([\gamma\cdot\tilde{s}, y]_\rho = [\tilde{s}, \rho(\gamma)^{-1}y]_\rho\). Thus, \(\varphi_{\gamma\cdot\tilde{s}} = \varphi_{\tilde{s}}\comp \rho(\gamma)^{-1}\). We define the map \(\Phi: P \times_G Y \to X\) by \[\Phi([[\tilde{s}, g]_\rho, y]_G) := [\tilde{s}, gy]_\rho.\]We verify this is well-defined. (i) Independence of the representative for \(p\in P\): Let \([\tilde{s}', g']_\rho = [\tilde{s}, g]_\rho\). This means there exists \(\gamma \in \pi_1(S,s_0)\) such that \(\tilde{s}' = \gamma \cdot \tilde{s}\) and \(g' = \rho(\gamma) g\). Then \[[\tilde{s}', g'y]_\rho = [\gamma\cdot\tilde{s}, (\rho(\gamma)g)y]_\rho.\]By the definition of the quotient \(X\), this equals \([\tilde{s}, \rho(\gamma)^{-1}((\rho(\gamma)g)y)]_\rho = [\tilde{s}, gy]_\rho\).

(ii) Independence of the representative for \([p, y]_G\): We use the standard associated bundle equivalence \([p\cdot h, h^{-1}y]_G = [p, y]_G\). If \(p=[\tilde{s}, g]_\rho\), then \(p\cdot h = [\tilde{s}, gh]_\rho\). \[\Phi([p\cdot h, h^{-1}y]_G) = \Phi([[\tilde{s}, gh]_\rho, h^{-1}y]_G) = [\tilde{s}, (gh)(h^{-1}y)]_\rho = [\tilde{s}, gy]_\rho.\]

The map \(\Phi\) is a smooth bundle isomorphism over \(S\). Let \(p=[\tilde{s}, g]_\rho\). We have \((\Phi \comp \phi_p)(y) = \Phi([p, y]_G) = [\tilde{s}, gy]_\rho\). On the other hand, \((\varphi_{\tilde{s}} \comp g)(y) = \varphi_{\tilde{s}}(gy) = [\tilde{s}, gy]_\rho\). Thus, \(\Phi \comp \phi_p = \varphi_{\tilde{s}} \comp g\). ◻

Proof. (2) \(\Rightarrow\) (1): Given a \(\rho\)-equivariant map \(h: \widetilde{S} \to \mathcal{H}_0\). We define \(\omega_s:=(\varphi_{\tilde{s}}^{-1})^* h(\tilde{s})\), where \(\tilde{s}\in \pi^{-1}(s)\). Let \(\tilde{s}'=\gamma\cdot\tilde{s}\), then \(\varphi_{\tilde{s}'}^\ast \omega_s = (\varphi_{\tilde{s}}\comp \rho(\gamma)^{-1})^\ast \omega_s = {(\rho(\gamma)^{-1})}^\ast (\varphi_{\tilde{s}}^\ast \omega_s)= {(\rho(\gamma)^{-1})}^\ast h(\tilde{s})\). By the equivariance of \(h\), this equals \(h(\tilde{s}')\). Thus we obtain a well-defined fiberwise Kähler metric \(\omega_{X/S}\). Let \(p=[\tilde{s}, g]_\rho \in P_s\). Using \(\phi_p = \varphi_{\tilde{s}} \comp g\) (by Lemma 34, where we omitted \(\Phi\)), we have \(\phi_p^* \omega_s = (\varphi_{\tilde{s}} \comp g)^* \omega_s = g^* (\varphi_{\tilde{s}}^* \omega_s) = g^* h(\tilde{s})\). Since \(h(\tilde{s}) \in \mathcal{H}_0\) and \(G\) preserves the class, \(g^* h(\tilde{s}) \in \mathcal{H}_0\).

(1) \(\Rightarrow\) (2): Given \(\omega_{X/S}\) \(G\)-modeled on \((Y, [\omega_Y])\). We define \(h(\tilde{s}) := \varphi_{\tilde{s}}^\ast \omega_s\). Consider \(p_e = [\tilde{s}, e]_\rho \in P_s\). Then \(\phi_{p_e} = \varphi_{\tilde{s}} \comp e = \varphi_{\tilde{s}}\). By Definition 18, \(\phi_{p_e}^* \omega_s \in [\omega_Y]\). Therefore, \(h(\tilde{s}) \in \mathcal{H}_0\). \(h\) is \(\rho\)-equivariant since \[h(\gamma\cdot \tilde{s}) = \varphi_{\gamma\cdot \tilde{s}}^\ast \omega_s = (\varphi_{\tilde{s}}\comp \rho(\gamma)^{-1})^\ast \omega_s = {(\rho(\gamma)^{-1})}^\ast (\varphi_{\tilde{s}}^\ast \omega_s) = {(\rho(\gamma)^{-1})}^\ast h(\tilde{s}). \qedhere\] ◻

Proof. We utilize the correspondence established in Lemma 35 and check that the condition of being \(G\)-modeled on \((Y, \omega_Y)\) is equivalent to the image of the corresponding map \(h\) being contained in \(G\cdot\omega_Y \subset \mathcal{H}_0\).

(2) \(\Rightarrow\) (1): Let \(p=[\tilde{s}, g]_\rho \in P_s\), \(\phi_p^* \omega_s = g^* h(\tilde{s})\in G\cdot\omega_Y\).

(1) \(\Rightarrow\) (2): Let \(p_e = [\tilde{s}, e]_\rho \in P_s\), \(\phi_{p_e}^* \omega_s \in G\cdot\omega_Y\). Since \(\varphi_{\tilde{s}} = \phi_{p_e}\), \(h(\tilde{s})=\phi_{p_e}^* \omega_s \in G\cdot\omega_Y\). ◻

Proof. Let \(P = \widetilde{S} \times_\rho G\) be the flat principal \(G\)-bundle associated to \(\rho\), equipped with the flat connection \(A\). By Lemma 34, we identify \(X\) with the associated bundle \(P \times_G Y\). Under this identification, the flat holomorphic connection \(\nabla^{1,0}_A\) on \(X\) is induced by \(A\). By Theorem 36, there exists a \(\rho\)-equivariant harmonic map \(h: \widetilde{S} \to G/K\), where \(K=\Stab_G(\omega_Y)\). This map determines a reduction of the structure group of \(P\) to \(K\), denoted by \(P_K\). By Lemma 29, this reduction induces the \(G\)-harmonic fiberwise Kähler metric \(\omega_{X/S}\) on \(X\).

With respect to the reduction \(\iota_K:P_K\to P\), the flat connection \(A\) decomposes as \(\iota_K^* A = A_K + \psi\), where \(A_K\) is a connection on \(P_K\) and \(\psi \in A^1(S, P_K\times_K \mathrm{i}\mathfrak{k})\). The \(G\)-Higgs bundle \((\mathcal{E}_G, \varphi)\) is defined by the principal \(\bar{\partial}\)-operator determined by \(A_K\) (extended to \(P\)) and the Higgs field \(\varphi = \psi^{1,0}\) (see [9] for \(G=\GL(n,\CC)\) and [31] for general reductive groups). We now compute the operators defined by the Simpson mechanism in Section 5.2. It suffices to work locally as in 32 , where we may write \(A=A^{1,0}+A^{0,1}\) with \(A^{1,0}\in A^{1,0}(U,\mathfrak{g})\), \(A^{0,1}\in A^{0,1}(U,\mathfrak{g})\). Let \(A_{\mathrm{Ch}}\) be the Chern connection on \(P\) determined by the holomorphic structure associated to \(A\) and the Hermitian structure \(P_K\). By Lemma 30, \(\partial_{\omega_{X/S}}^{\mathrm{Ch}}\) is induced by \(A_{\mathrm{Ch}}\). Then \(\theta_{\omega_{X/S}}\) is induced by \((A-A_{\mathrm{Ch}})^{1,0}/2\). Locally, \[\psi^{1,0}(\partial_i)=\varphi(\partial_i)=(\psi(\partial_{x^i})-\mathrm{i}\psi(\partial_{y^i}))/2,\quad \psi^{0,1}(\partial_{\bar{i}})=(\psi(\partial_{x^i})+\mathrm{i}\psi(\partial_{y^i}))/2=\bar{\varphi}_K(\partial_{\bar{i}}).\] Note that \(\psi(\partial_{x^i}),\psi(\partial_{y^i})\in \mathrm{i}\mathfrak{k}\). This implies \(A_{\mathrm{Ch}}^{1,0}=A_K^{1,0}-\varphi\), and then \((A-A_{\mathrm{Ch}})^{1,0}/2=\varphi\). By Lemma 33, the conjugate \(\bar{\theta}_{\omega_{X/S}}\) is induced by \(\bar{\varphi}_K=\psi^{0,1}\). Then \(\bar{\partial}_{\omega_{X/S}}=\bar{\partial}_\nabla-\bar{\theta}_{\omega_{X/S}}\) is induced by \(A^{0,1}-\psi^{0,1}=A_K^{0,1}\). Therefore \((\bar{\partial}_{\omega_{X/S}},\theta_{\omega_{X/S}})\) is induced by \((\mathcal{E}_G, \varphi)\), which is a nonlinear Higgs bundle.

Finally, we address uniqueness. Let \(h'\) be another equivariant harmonic map. By the uniqueness (Remark 37), there exists \(g \in Z_G(\im \rho)\) such that \(h' = g \cdot h\). This element \(g\) defines a gauge transformation \(\Phi_g: P \to P\) by \(\Phi_g([\tilde{s}, a]_\rho) := [\tilde{s}, g \cdot a]_\rho\), which is well-defined since \(g\) commutes with \(\rho(\gamma)\). Since \(g\) is constant, \(\Phi_g\) preserves the flat connection \(A\). The harmonic map \(h' = g \cdot h\) defines a reduction \(P'_K\) which is precisely the image of \(P_K\) under \(\Phi_g\). Then \(\Phi_g\) maps the decomposition of \(A\) (relative to \(h\)) to the decomposition of \(A\) (relative to \(h'\)). Thus, \(\Phi_g\) induces an isomorphism of the corresponding \(G\)-Higgs bundles. Note that any gauge transform \(\Phi\) of \(P\) induces an automorphism \(\Psi_\Phi\) of the associated bundle \(f: X = P \times_G Y \to S\) via \(\Psi_\Phi([p, y]) := [\Phi(p), y]\). Since \(\Phi_g\) is an isomorphism of Higgs bundles, \(\Psi_{\Phi_g}\) is an isomorphism of the induced nonlinear Higgs bundles.

Let \(\omega_{Y}'\) be another Kähler metric on \(Y\) such that \(K'= \Stab_G(\omega_{Y}')\) is also a compact real form of \(G\). There exists \(u \in G\) such that \(K' = u K u^{-1}\). This conjugation induces a \(G\)-equivariant isometry of symmetric spaces \(\iota: G/K \to G/K'\) by \(x K \mapsto x u^{-1} K'\). Then \(h' := \iota \comp h\) is a \(\rho\)-equivariant harmonic map \(\widetilde{S}\to G/K'\). Let \(\sigma\) be a local frame corresponding to the Hermitian structure induced by \(h\). Then \(\sigma' = \sigma u^{-1}\) corresponds to \(h'\). The flat connection in these frames are expressed by \(A\in A^1(U,\mathfrak{g})\) and \(A'= u A u^{-1}\) respectively. Correspondingly, we have the splitting \[A'=A_K'+\psi', \text{ where } A_K'=u A_K u^{-1},~\psi'=u \psi u^{-1}.\] Therefore, the Higgs bundles induced by \(h\) and \(h'\) are identical.

The Higgs bundle structure \((f, \bar{\partial}_{\omega_{X/S}}, \theta_{\omega_{X/S}})\) is independent of \(\omega_S\) since the harmonicity implies the pluriharmonicity by the Siu–Sampson theorem [31] and the pluriharmonicity is independent of \(\omega_S\). ◻

Proof. Let \(F: \mathcal{X}_1 \to \mathcal{X}_2\) be a morphism in \(\mathbf{RFB}(S)\), induced by the \(\alpha\)-equivariant holomorphic map \(\phi: Y_1 \to Y_2\). Let \(h_1: \widetilde{S} \to G_1/K_1\) be the harmonic map chosen for \(\mathcal{X}_1\). The condition \(\alpha(K_1) \subset K_2\) implies that \(\alpha\) induces a smooth map \(\tilde{\alpha}: G_1/K_1 \to G_2/K_2\) defined by \(gK_1 \mapsto \alpha(g)K_2\). Since \(\alpha\) maps the Cartan involution of \(G_1\) to that of \(G_2\), \(\tilde{\alpha}\) is a totally geodesic map between the symmetric spaces.

Consider the map \(h'_2 := \tilde{\alpha} \comp h_1: \widetilde{S} \to G_2/K_2\). Since the composition of a harmonic map with a totally geodesic map is harmonic, \(h'_2\) is harmonic. Moreover, it is equivariant with respect to \(\rho_2 = \alpha \comp \rho_1\) since \[h'_2(\gamma \cdot \tilde{s}) = \tilde{\alpha}(h_1(\gamma \cdot \tilde{s})) = \tilde{\alpha}(\rho_1(\gamma) \cdot h_1(\tilde{s})) = \alpha(\rho_1(\gamma)) \cdot \tilde{\alpha}(h_1(\tilde{s})) = \rho_2(\gamma) \cdot h'_2(\tilde{s}).\] Thus, \(h'_2\) corresponds to a \(G\)-harmonic metric for \(\mathcal{X}_2\). By the uniqueness of equivariant harmonic maps, \(h'_2=g\cdot h_2\), where \(g \in Z_{G_2}(\im\rho_2)\) and \(h_2\) is the chosen harmonic map for \(\mathcal{X}_2\).

Since \(\phi\) is \(\alpha\)-equivariant, we have \(\phi_* (\tau_{0,Y_1}(\xi)) = \tau_{0,Y_2}(\alpha_*(\xi))\) for any \(\xi \in \mathfrak{g}_1\). \(F\) is induced by \(\phi\), so it defines a morphism of Higgs structures from \(\mathsf{H}(\mathcal{X}_1)\) to the structure defined by \(h'_2\). Composing with the isomorphism \(\Psi_g^{-1}\), we obtain that \(\widetilde{F} := \Psi_g^{-1} \comp F\) is a morphism from \(\mathsf{H}(\mathcal{X}_1)\) to \(\mathsf{H}(\mathcal{X}_2)\). We define \(\mathsf{H}(F) = \widetilde{F}\).

The functor \(\mathsf{H}\) assigns to a bundle map \(F\) the map \(\Psi_g^{-1} \comp F\). Since \(\Psi_g\) is a smooth automorphism of the bundle, the mapping \(F \mapsto \Psi_g^{-1} \comp F\) is injective. If \(\mathsf{H}(F) = \mathsf{H}(G)\), then \(\Psi_{g_F}^{-1} \comp F = \Psi_{g_G}^{-1} \comp G\). Since \(F\) and \(G\) map to the same flat bundle with harmonic metric \(h'_2\) (determined by \(\alpha\) and is independent of \(\phi\)), the gauge adjustments \(g_F\) and \(g_G\) are identical. Thus \(F=G\), and the functor is faithful. ◻

Proof. It suffices to show the action \(G \curvearrowright \mathcal{H}_0\) is proper. It is known that the action \(\Diff(Y) \curvearrowright \mathcal{R}(Y)\) is proper, where \(\mathcal{R}(Y)\) is the space of Riemannian metrics (see e.g. [39]). By [39], we only need to show that for any \(\omega\in \mathcal{H}_0\), if a sequence \(\{g_n\}\) in \(G\) is such that \(\omega_n = g_n^*\omega\) converges smoothly in \(\mathcal{H}_0\) to \(\omega\), then \(\{g_n\}\) has a convergent subsequence in \(G\). By [39], there is a subsequence (still denoted by \(\{g_n\}\)) converging in \(C^\infty\) to \(g\in \Diff(Y)\). The maps \(g_n\in \Aut_0(Y)\) are holomorphic, so the limit \(g\) is also holomorphic. Since \(g\in \Diff(Y)\), \(g^{-1}\) is also holomorphic. Therefore, \(g\in \Aut_0(Y)\). Since \(G\) is a closed subgroup of \(\Aut_0(Y)\), and \(g_n \in G\) converges to \(g\) in \(\Aut_0(Y)\), the limit \(g\) must belong to \(G\). ◻

Proof. By the assumption, \(G\cdot\omega_Y\cong G/K\) is a Riemannian symmetric space. Let \(\mathfrak{a}\subset \tilde{\mathfrak{g}}\) be the abelian ideal of harmonic vector fields, where \(\widetilde{G}=\Aut_0(Y)\) and \(\tilde{\mathfrak{g}}= H^0(Y,TY)\) is its Lie algebra (the identification is provided by \(\tau_0\) in Lemma 11). Fix \(\omega\in G\cdot\omega_Y\), and let \(K_\omega=\Stab_G(\omega)\), which is also a compact real form of \(G\). Let \(\mathfrak{k}_\omega\) be the Lie algebra of \(K_\omega\). By [27], \(\mathfrak{g}=\mathfrak{g}_1\oplus \mathfrak{g}_{\mathfrak{a}}\), where \(\mathfrak{g}_{\mathfrak{a}}=\mathfrak{g}\cap\mathfrak{a}\), \(\mathfrak{g}_1=\mathfrak{g}\cap \{\grad^{1,0}_\omega f\mid f\in C^{\infty}(Y,\CC)\}=\mathfrak{k}_1\oplus \mathrm{i}\mathfrak{k}_1\), \(\mathfrak{k}_1=\mathfrak{g}\cap \{\grad^{1,0}_\omega f\mid f\in \mathrm{i}C^{\infty}(Y,\RR)\}=\mathfrak{k}_\omega\cap\mathfrak{g}_1\). We have \(T_{\omega} (G\cdot\omega_Y)=\{L_{v_\RR}\omega| v\in \mathfrak{g}\}\), where \(L_{v_\RR}\) denotes the Lie derivative. By [27], for \(v\in\mathfrak{g}\), \(L_{v_\RR}\omega_Y=0\) if and only if \(v\in \mathfrak{k}_1\oplus \mathfrak{g}_{\mathfrak{a}}\). Therefore, \(T_{\omega} (G\cdot\omega_Y)=\mathrm{i}\mathfrak{k}_1\). Since \(\dim G=2\dim (G\cdot\omega_Y)\), we have \(\mathfrak{g}_{\mathfrak{a}}=\{0\}\), and then \(\mathfrak{g}=\mathfrak{g}_1\), \(\mathfrak{k}=\mathfrak{k}_1\). Define \(\mu: Y \to \mathfrak{k}^*\) by \(\langle \mu(y), \xi \rangle := f_\xi(y)\), where \(f_\xi\in C^{\infty}(Y,\RR)_0\) is the unique mean-zero function such that \(\tau_0(\xi)=\grad^{1,0}_{\omega_Y}(\mathrm{i}f_\xi)\). Then \(\mu\) is a moment map for \(K\).

Since \(G\cdot \omega_Y\) is Riemannian symmetric, every geodesic \(\gamma(t)\) in \(G\cdot \omega_Y\) through \(\omega\) has the form \(\gamma(t)=(\exp(t v_\RR))^*\omega\) for some nonzero \(v\in \mathrm{i}\mathfrak{k}\), which is a geodesic in \(\mathcal{H}_0\) by [40]. Therefore, \(G\cdot\omega_Y\) is totally geodesic. ◻

Proof. We have \(\mathfrak{k}= \{ v \in \mathfrak{g} \mid L_{v_\RR}\omega_Y = 0 \}\). We aim to show that \(\mathfrak{g}= \mathfrak{k} \oplus \mathrm{i}\mathfrak{k}\). As above, we decompose \(\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_{\mathfrak{a}}\). Elements in \(\mathfrak{a}\) are Killing. Thus, \(\mathfrak{g}_{\mathfrak{a}} \subset \mathfrak{k}\). Let \(v \in \mathfrak{g}\), decomposed as \(v = v_1 + v_{\mathfrak{a}}\). The corresponding tangent vector to the orbit is \(V=L_{v_\RR}\omega_Y= L_{(v_1)_\RR}\omega_Y\). We can write \(v_1=\grad^{1,0}_{\omega_Y}f_1+\grad^{1,0}_{\omega_Y}(\mathrm{i}f_2)=:u_1+u_2\). Then by [27], \(V=L_{(u_1)_\RR}\omega_Y\). By [40], the unique geodesic in \(\mathcal{H}_0\) starting at \(\omega_Y\) with velocity \(V\) is \(\gamma(t) = (\exp(t(u_1)_\RR))^* \omega_Y\). By hypothesis, \(G\cdot\omega_Y\) is totally geodesic, so \(\gamma(t) \in G\cdot\omega\). Since \(G\) is connected, the infinitesimal generator \(u_1\) must belong to \(\mathfrak{g}\), and \(u_2 = v_1 - u_1\in \mathfrak{g}_1\). Define \(\mathfrak{k}_1=\mathfrak{g}_1\cap \mathfrak{k}\). Then \(\mathfrak{g}_1 = \mathfrak{k}_1 \oplus \mathrm{i}\mathfrak{k}_1\), and we have \(\mathfrak{k} = \mathfrak{k}_1 \oplus \mathfrak{g}_{\mathfrak{a}}\). Therefore, \(\dim_{\CC}(\mathfrak{g}_1) = \dim_{\RR}(\mathfrak{k}_1)\), \(\dim_{\RR}(\mathfrak{k})= \dim_{\RR}(\mathfrak{k}_1) + \dim_{\RR}(\mathfrak{g}_{\mathfrak{a}})\geq \dim_\CC(\mathfrak{g})\). Since \(K\) is a compact subgroup of \(G\), we have \(\dim_\RR(\mathfrak{k})\leq \dim_\CC(\mathfrak{g})\). Therefore, \(\dim_\RR(\mathfrak{k})= \dim_\CC(\mathfrak{g})\) and \(\mathfrak{g}_{\mathfrak{a}} = \{0\}\). This yields \(\mathfrak{g} = \mathfrak{g}_1\) and \(\mathfrak{k} = \mathfrak{k}_1\). The decomposition \(\mathfrak{g} = \mathfrak{k} \oplus i\mathfrak{k}\) holds, proving that \(K\) is a compact real form of \(G\). ◻

Proof. Suppose we are given a complex bundle isomorphism \(\beta_Y: TY_A \to TY_B\). We define \(\beta_Y^\RR: TY^\RR \to TY^\RR\) by \(\beta_Y^\RR(u) := \beta_Y(v) + \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\beta_Y(v)} \endgroup\), where \(u=v+\bar{v}\) with \(v\in TY_A\). Then we have \[\begin{align} \beta_Y^\RR (J_A u) &= \beta_Y(\mathrm{i}v) + \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\beta_Y({\mathrm{i}} v)} \endgroup = \mathrm{i}\beta_Y(v) - \mathrm{i}\overline{\beta_Y(v)}\\ &=J_B(\beta_Y(v) + \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\beta_Y(v)} \endgroup )=J_B(\beta_Y^\RR(u)). \end{align}\] Conversely, let \(\beta_Y^\RR: TY^\RR \to TY^\RR\) be a real isomorphism such that \(\beta_Y^\RR \comp J_A = J_B \comp \beta_Y^\RR\). We extend \(\beta_Y^\RR\) complex-linearly to \(\beta_Y^\CC: TY^\CC \to TY^\CC\). For any \(v \in TY_A\), \[J_B(\beta_Y^\CC (v)) = \beta_Y^\CC(J_A v) = \beta_Y^\CC(\mathrm{i}v) = \mathrm{i}\beta_Y^\CC(v).\] This implies that \(\beta_Y :=\beta_Y^{\mathbb{C}} \big|_{TY_A} : TY_A \to TY_B\) is a complex bundle isomorphism. ◻

Proof. We prove the first statement. Since \(v\in \mathfrak{k}^\CC\), we have \(v=u+\mathrm{i}w\) and \(\bar{v}_{\mathfrak{k}}=u-\mathrm{i}w\) for \(u,w\in \mathfrak{k}\). Since \(u,w\) are Hamiltonian, we have \[\iota_v\omega_Y=\iota_{u+\mathrm{i}w}\omega_Y=\bar{\partial}(f_u+\mathrm{i}f_w)=:\bar{\partial}(f_v),\] for some real functions \(f_u\) and \(f_w\). Then we have \(\iota_{\bar{v}_{\mathfrak{k}}}\omega_Y=\bar{\partial} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{f_v} \endgroup\).

Now we prove the second statement. We have \(\iota_v (\omega_{J_A} - \mathrm{i}\omega_{J_C}) = \partial f_v\). Taking the complex conjugate, \[\iota_{\bar{v}} (\omega_{J_A} + \mathrm{i}\omega_{J_C}) = \bar{\partial} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{f_v} \endgroup = \mathrm{d} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{f_v} \endgroup .\] Since \(\Omega_{J_{B}}\) has type \((2,0)\) and \(\bar{v}\) has type \((0,1)\) with respect to \(J_B\), we have \(\iota_{\bar{v}}\Omega_{J_B}=0\), implying \(\iota_{\bar{v}} \omega_{J_A} = \mathrm{i}\iota_{\bar{v}} \omega_{J_C}\). Then \(\iota_{\bar{v}} \omega_{J_C} = -\frac{\mathrm{i}}{2} \mathrm{d} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{f_v} \endgroup\). On the other hand, \[\iota_{\mathrm{i}J_A\bar{v}} \omega_{J_B} (\cdot) = \mathrm{i}g(J_B(J_A\bar{v}), \cdot)=-\mathrm{i}\iota_{\bar{v}}\omega_{J_C}=-\tfrac{1}{2} \mathrm{d} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{f_v} \endgroup .\] Therefore, \(\tilde{v}=\mathrm{i}J_A(\bar{v})\). ◻

Proof. Since \(TY_A\) and \(TY_B\) are isomorphic Hermitian bundles, we can choose an isometry \(\beta_0\) satisfying \(\beta_0 \comp J_A = J_B \comp \beta_0\). Define a vector field \(v' := \beta_0(v)\). Since \(\beta_0\) is an isometry, we have \(|v'|_{g_Y} = |v|_{g_Y} = |w|_{g_Y}\). Thus, \(v'\) and \(w\) are both sections of the sphere bundle \(S(TY) \subset TY\) of the same radius. Since \(v\) is homotopic to \(w\) and \(\beta_0\) is a bundle isomorphism, \(v'\) is homotopic to \(w\). Since the unitary group acts transitively on the sphere, there is a fiber bundle \(\pi_{v'}:\Aut(TY, J_B, g_Y)\to S(TY)\) with typical fiber \(U(m-1)\), where \(\Aut(TY, J_B, g_Y)\) is the bundle of unitary transformations and \(\pi_{v'}(U)=U(v')\). By the homotopy lifting property, there exists \(U\in \Aut(TY, J_B, g_Y)\) such that \(U(v')=w\). Define \(\beta_Y^\RR := U \comp \beta_0\), which is an isometry since it is a composition of isometries. We have \(\beta_Y^\RR(v) = U(\beta_0(v)) = U(v') = w\) and \(\beta_Y^\RR J_A = U \beta_0 J_A = U J_B \beta_0 = J_B U \beta_0 = J_B \beta_Y^\RR\). ◻

Proof. Let \(M = \nabla_{g_Y} v\). The \(J_A\)-holomorphicity of \(v\) implies \(M J_A = J_A M\). The \(J_B\)-holomorphicity of \(w\) implies \((\nabla_{g_Y} w) J_B = J_B (\nabla_{g_Y} w)\). Since \(\beta_Y^\RR\) is parallel, \(\nabla_{g_Y} w = \nabla_{g_Y}(\beta_Y^\RR v) = \beta_Y^\RR M\). We have \[(\beta_Y^\RR M) J_B = J_B (\beta_Y^\RR M)=\beta_Y^\RR J_A M.\] Thus \(M J_B = J_A M=M J_A\), and then \(M=\nabla_{g_Y} v = 0\). ◻

Proof. Let \(\beta_Y^\RR = a_0 \id + a_1 J_A + a_2 J_B + a_3 J_C\). Imposing \(\beta_Y^\RR\comp J_A = J_B\comp \beta_Y^\RR\) leads to the constraints \(a_0 = a_3\) and \(a_1 = a_2\). Next, we compute \[\begin{align} (\beta_Y^\RR)^* \beta_Y^\RR &= \bigl(c_1(\id - J_C) - c_2(J_A + J_B)\bigr) \bigl(c_1(\id + J_C) + c_2(J_A + J_B)\bigr)\\ &=2(c_1^2 + c_2^2)\id. \end{align}\] Setting this to \(\id\) requires \(2(c_1^2 + c_2^2) = 1\). ◻

Proof. A function \(f\) is holomorphic if \({J_{\Pi}^A}^*\mathrm{d}f = \mathrm{i}\mathrm{d}f\). By the definition, we have \({J_\Pi^A}^* \mathrm{d}q_x = \Pi_y \mathrm{d}p_y\), \({J_\Pi^A}^* \mathrm{d}p_y = -\Pi_y^{-1} \mathrm{d}q_x\), \({J_\Pi^A}^* \mathrm{d}q_y = \Pi_y \mathrm{d}p_x\), \({J_\Pi^A}^* \mathrm{d}p_x = -\Pi_y^{-1} \mathrm{d}q_y\). We have \(\mathrm{d}\xi = \Pi_y^{-1} \mathrm{d}q_y - \mathrm{i}\mathrm{d}p_x\), and \[{J_\Pi^A}^*\mathrm{d}\xi=\Pi_y^{-1} \Pi_y \mathrm{d}p_x+\mathrm{i}\Pi_y^{-1}\mathrm{d}q_y=\mathrm{i}\mathrm{d}\xi.\] Thus \(\xi\) is holomorphic. For \(\eta\), we have \[\begin{align} {J_\Pi^A}^* \mathrm{d}\Re(\eta) &= {J_\Pi^A}^* (\mathrm{d}q_x - \Pi_x \Pi_y^{-1} \mathrm{d}q_y)= \Pi_y \mathrm{d}p_y - \Pi_x \Pi_y^{-1} (\Pi_y \mathrm{d}p_x) = \Pi_y \mathrm{d}p_y - \Pi_x \mathrm{d}p_x. \\ \mathrm{i}{J_\Pi^A}^*(\mathrm{d}\Im(\eta)) &= \mathrm{i}{J_\Pi^A}^*(\Pi_x \mathrm{d}p_x - \Pi_y \mathrm{d}p_y)=\mathrm{i}(\Pi_x (-\Pi_y^{-1} \mathrm{d}q_y) - \Pi_y (-\Pi_y^{-1} \mathrm{d}q_x)) =\mathrm{i}(\mathrm{d}q_x-\Pi_x \Pi_y^{-1} \mathrm{d}q_y). \end{align}\] Therefore, \({J_\Pi^A}^* (\mathrm{d}\Re(\eta) + \mathrm{i}\mathrm{d}\Im(\eta)) = -\mathrm{d}\Im(\eta) + \mathrm{i}\mathrm{d}\Re(\eta) = \mathrm{i}(\mathrm{d}\Re(\eta) + \mathrm{i}\mathrm{d}\Im(\eta))\). ◻

Proof. A vector field is horizontal with respect to the Gauss-Manin connection if \(\xi\) and \(\eta\) are constant along its integral curve. We compute the induced variations \(\dot{q}\) and \(\dot{p}\) by differentiating 75 with respect to \(\Pi\) in the direction \(V_{ij}\). Note that \(\dot{\Pi} = V_{ij}\) and \(\dot{\overline{\Pi}} = 0\). We have \[\dot{q} = \overset{\cdot}{\wideparen{\Re(\eta)}} + \dot{\Pi} \Re(\xi) + \Pi \overset{\cdot}{\wideparen{\Re(\xi)}} = \dot{\Pi} \Re(\xi)= V_{ij} \Pi_y^{-1} q_y.\] For the conjugate coordinate \(\bar{q} = \Re(\eta) + \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Pi} \endgroup \Re(\xi)\), the derivative is \[\dot{\bar{q}} = \dot{\overline{\Pi}} \Re(\xi) = 0.\] By 75 , the derivative of \(\Pi_y p\) vanishes, so \[\dot{\Pi}_y p + \Pi_y \dot{p} = 0.\] Note that \(\dot{\Pi}_y = \frac{1}{2\mathrm{i}} (\dot{\Pi}- \dot{\overline{\Pi}}) = \frac{1}{2\mathrm{i}} V_{ij}\). Then \[\dot{p} = - \Pi_y^{-1} \Bigl( \frac{1}{2\mathrm{i}} V_{ij} p \Bigr) = \tfrac{\mathrm{i}}{2} \Pi_y^{-1} V_{ij} p.\] Since \(p + \bar{p} = -2\Im(\xi)\) is constant, we have \[\dot{\bar{p}} = -\dot{p} = - \tfrac{\mathrm{i}}{2} \Pi_y^{-1} V_{ij} p.\] Therefore ?? follows. By taking its conjugate we obtain \(\nabla^{0,1}\), which induces \(\bar{\partial}_{B,0}\). The conjugate of the last term in ?? is \((\bar{\partial}_{B,0}-\bar{\partial}_B)(\partial_{ \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{ij} \endgroup })\), and the second statement follows. ◻

Proof. The two holomorphic vector fields on \(X_\Pi\) are \[\theta(\partial_{i_1 j_1}) = f^a(p) \partial_{q^a} \quad \text{and}\quad \theta(\partial_{i_2 j_2}) = g^c(p) \partial_{q^c},\] where \(f^a(p) = -\frac{\mathrm{i}}{2} (V_{i_1 j_1})_{ab} p^b\) and \(g^c(p) = -\frac{\mathrm{i}}{2} (V_{i_2 j_2})_{cd} p^d\). Then \[[\theta(\partial_{i_1 j_1}), \theta(\partial_{i_2 j_2})] = (f^a(p) \partial_{q^a} g^c(p) - g^a(p) \partial_{q^a} f^c(p))\partial_{q^c}=0,\] since \(f^a(p)\) and \(g^c(p)\) are independent of \(q\). ◻

Proof. By the symmetry of \(V_{ij}\), we have \[\mathrm{d}H_{ij}=\tfrac{\mathrm{i}}{2} (V_{ij} p) \cdot \mathrm{d}p.\] On the other hand, \[\begin{align} \iota_{\theta(\partial_{ij})} (\mathrm{d}p^\alpha \wedge \mathrm{d}q^\alpha) &= (\iota_{\theta(\partial_{ij})} \mathrm{d}p^\alpha) \wedge \mathrm{d}q^\alpha - \mathrm{d}p^\alpha \wedge (\iota_{\theta(\partial_{ij})} \mathrm{d}q^\alpha) \\ &= 0 - \mathrm{d}p^\alpha \wedge \bigl( -\tfrac{\mathrm{i}}{2} (V_{ij})_{ab} p^b \delta_{\alpha a} \bigr) \\ &= \tfrac{\mathrm{i}}{2} (V_{ij})_{a b} p^b \mathrm{d}p^a=\mathrm{d}H_{ij}. \qedhere \end{align}\] ◻

Proof. Let \(v\) be the vector field in 76 , it suffices to show \[\mathrm{i}J_\Pi^A(\bar{v}_{J_\Pi^B})=\tfrac{\mathrm{i}}{2}\Pi_y^{-1}V_{ij}\bar{p}\cdot\partial_{p}.\] First we have \[\bar{v}_{J_\Pi^B} = \overline{-\tfrac{\mathrm{i}}{2} (V_{ij} p)^a} \partial_{\bar{q}^a} = \tfrac{\mathrm{i}}{2} (V_{ij} \bar{p})^a \partial_{\bar{q}^a}.\] Using \(J_\Pi^A (\partial_{q_x}) = -\Pi_y^{-1} \partial_{p_y}\) and \(J_\Pi^A (\partial_{q_y}) = -\Pi_y^{-1} \partial_{p_x}\), we have \[\begin{align} J_\Pi^A (\partial_{\bar{q}^a}) &= \tfrac{1}{2} \bigl( J_\Pi^A(\partial_{q_x^a}) + \mathrm{i}J_\Pi^A(\partial_{q_y^a}) \bigr) \\ &= \tfrac{1}{2} \bigl( -(\Pi_y^{-1})^{ab} \partial_{p_y^b} - \mathrm{i}(\Pi_y^{-1})^{ab} \partial_{p_x^b} \bigr) \\ &=-\mathrm{i}(\Pi_y^{-1})^{ab} \partial_{p^b}. \end{align}\] Then we have \[\begin{align} \mathrm{i}J_\Pi^A (\bar{v}_{J_\Pi^B}) &= \mathrm{i}\bigl( \tfrac{\mathrm{i}}{2} (V_{ij} \bar{p})^a \bigr) J_\Pi^A (\partial_{\bar{q}^a}) \\ &= -\tfrac{1}{2}(V_{ij} \bar{p})^a \bigl( -\mathrm{i}(\Pi_y^{-1})^{ab} \partial_{p^b} \bigr) \\ &= \tfrac{\mathrm{i}}{2} (\Pi_y^{-1})^{ba} (V_{ij} \bar{p})^a \partial_{p^b}=\tfrac{\mathrm{i}}{2} \Pi_y^{-1} V_{ij} \bar{p}\cdot \partial_p, \end{align}\] where we used the symmetry \((\Pi_y^{-1})^{ba} = (\Pi_y^{-1})^{ab}\). ◻

Proof. It suffices to show that \(\mathrm{i}\partial_{X_\Pi^A}\bar{\partial}_{X_\Pi^A}\phi=\omega_{J_\Pi^A}\). In the following, we omit the subscript \(X_\Pi^A\). It is equivalent to show \(\frac{1}{2} \mathrm{d}\mathrm{d}^c \phi = \omega_{J_\Pi^A}\), since \(\mathrm{d}\mathrm{d}^c = 2 \mathrm{i}\partial\bar{\partial}\) for \(\mathrm{d}^c = -{J_\Pi^A}^* \mathrm{d}\). Let \(p^\alpha = p_x^\alpha + \mathrm{i}p_y^\alpha\), then \[\phi = (\Pi_y)_{\alpha\beta} (p_x^\alpha + \mathrm{i}p_y^\alpha) (p_x^\beta - \mathrm{i}p_y^\beta) = (\Pi_y)_{\alpha\beta} (p_x^\alpha p_x^\beta + p_y^\alpha p_y^\beta),\] where we used the symmetry of \(\Pi_y\). We compute \[\begin{align} \mathrm{d}^c \phi &= -{J_\Pi^A}^* \bigl(2 (\Pi_y)_{\alpha\beta} p_x^\alpha \mathrm{d}p_x^\beta + 2 (\Pi_y)_{\alpha\beta} p_y^\alpha \mathrm{d}p_y^\beta\bigr)\\ &=- \bigl( 2 (\Pi_y)_{\alpha\beta} p_x^\alpha ( -(\Pi_y^{-1})^{\beta a} \mathrm{d}q_y^a ) + 2 (\Pi_y)_{\alpha\beta} p_y^\alpha ( -(\Pi_y^{-1})^{\beta a} \mathrm{d}q_x^a) \bigr)\\ &=2 p_x^a \mathrm{d}q_y^a + 2 p_y^a \mathrm{d}q_x^a. \end{align}\] Taking the exterior derivative, \[\mathrm{d}(\mathrm{d}^c \phi) = 2 \mathrm{d}p_x^a \wedge \mathrm{d}q_y^a + 2 \mathrm{d}p_y^a \wedge \mathrm{d}q_x^a=2\omega_{J_\Pi^A}.\] The conclusion follows. ◻

Proof. We work in the holomorphic coordinates \((\Pi_{ij},\xi^\alpha, \eta^a)\), where the potential \(\phi\) depends only on the imaginary parts \(\xi_y = \Im(\xi)\) and \(\eta_y = \Im(\eta)\). The connection coefficients \(\Gamma^{\mathsf{a}}_{ij}\) (where \(\mathsf{a}=a\) or \(\alpha\)) are determined by \[g_{\mathsf{a}\bar{\mathsf{b}}} \Gamma_{ij}^{\mathsf{b}} = - g_{ij,\bar{\mathsf{b}}}, \quad \text{where } g_{\mathsf{a}\bar{\mathsf{b}}} = \partial_{\mathsf{a}} \partial_{\bar{\mathsf{b}}} \phi.\] Recall that \(\phi = p^\dagger \Pi_y p\). Using 75 , \(p_x = -\xi_y\) and \(p_y = -\Pi_y^{-1}(\eta_y + \Pi_x \xi_y)\), then \[\label{eq:phi95xi95eta} \phi(\Pi, \xi_y, \eta_y) = \xi_y^{\text{T}} \Pi_y \xi_y + (\eta_y + \Pi_x \xi_y)^{\text{T}} \Pi_y^{-1} (\eta_y + \Pi_x \xi_y).\tag{78}\] We have \(\frac{\partial \phi}{\partial \bar{\xi}}=\frac{\mathrm{i}}{2}\frac{\partial \phi}{\partial \xi_y}\), \(\frac{\partial \phi}{\partial \xi}=-\frac{\mathrm{i}}{2}\frac{\partial \phi}{\partial \xi_y}\), and similarly for \(\eta\). The fiber coefficient matrix of \(\omega\) is \[(g_{\mathsf{a}\bar{\mathsf{b}}}) = \tfrac{1}{4} \operatorname{Hess}_{\xi_y,\eta_y}(\phi)= \tfrac{1}{2} \begin{pmatrix} \Pi_y + \Pi_x^{\text{T}} \Pi_y^{-1} \Pi_x & \Pi_x^{\text{T}} \Pi_y^{-1} \\ \Pi_y^{-1} \Pi_x & \Pi_y^{-1} \end{pmatrix}.\] Its inverse is \[\label{eq:G95inv95final} (g^{\bar{\mathsf{b}}\mathsf{a}}) = 2 \begin{pmatrix} \Pi_y^{-1} & -\Pi_y^{-1} \Pi_x \\ -\Pi_x \Pi_y^{-1} & \Pi_y + \Pi_x \Pi_y^{-1} \Pi_x \end{pmatrix}.\tag{79}\]

We compute \(g_{ij,\bar{\mathsf{b}}} = \partial_{ij} (\partial_{\bar{\mathsf{b}}} \phi)\). Recall \(V_{ij} = \dot{\Pi}\), then \(\dot{\Pi}_x = \frac{1}{2}(\dot{\Pi}+\dot{\overline{\Pi}}) =\frac{1}{2}V_{ij}\) and \(\dot{\Pi}_y = -\frac{\mathrm{i}}{2}V_{ij}\). Note that \(\Pi_y\) is symmetric, \(\partial_{\eta_y} \phi=(\partial p_y/\partial \eta_y)\partial_{p_y} \phi = -\Pi_y^{-1}(2\Pi_y p_y)=-2p_y\). Then \[\begin{align} g_{ij,\bar{\eta}}&= \tfrac{\mathrm{i}}{2}\partial_{ij}(-2p_y)=\mathrm{i}\partial_{ij}\bigl(\Pi_y^{-1}(\eta_y + \Pi_x \xi_y)\bigr)\\ &=\mathrm{i}\bigl(\tfrac{\mathrm{i}}{2} \Pi_y^{-1} V_{ij} \Pi_y^{-1} (\eta_y + \Pi_x \xi_y) + \Pi_y^{-1} \big(\tfrac{1}{2} V_{ij} \xi_y\big) \bigr)\\ &=\mathrm{i}\bigl(\tfrac{\mathrm{i}}{2} \Pi_y^{-1} V_{ij} (-p_y) + \tfrac{1}{2} \Pi_y^{-1} V_{ij} (-p_x) \bigr)\\ &=-\tfrac{\mathrm{i}}{2} \Pi_y^{-1} V_{ij} p. \end{align}\] Similarly, \(\partial_{\xi_y} \phi = 2 \Pi_y \xi_y - 2 \Pi_x p_y\), and using \(\partial_{ij} p_y=(1/2)\Pi_y^{-1}V_{ij}p\) computed above, \[\begin{align} g_{ij,\bar{\xi}}&=\partial_{ij} \partial_{\bar{\xi}}\phi =\mathrm{i}\bigl(-\tfrac{\mathrm{i}}{2} V_{ij} \xi_y - \tfrac{1}{2} V_{ij} p_y - \Pi_x (\partial_{ij} p_y)\bigr) \\ &=\tfrac{\mathrm{i}}{2}( V_{ij} (\mathrm{i}p_x - p_y) - \Pi_x \Pi_y^{-1} V_{ij} p) \\ &= \tfrac{\mathrm{i}}{2} (\mathrm{i}V_{ij} p - \Pi_x \Pi_y^{-1} V_{ij} p). \end{align}\]

Therefore, the coefficients of the symplectic connection are \[\begin{align} \begin{pmatrix} \Gamma_{ij}^\xi \\ \Gamma_{ij}^\eta \end{pmatrix} &= -2 \begin{pmatrix} \Pi_y^{-1} & -\Pi_y^{-1} \Pi_x \\ -\Pi_x \Pi_y^{-1} & \Pi_y + \Pi_x \Pi_y^{-1} \Pi_x \end{pmatrix} \begin{pmatrix} \frac{\mathrm{i}}{2}(\mathrm{i}V_{ij} p - \Pi_x \Pi_y^{-1} V_{ij} p) \\ -\frac{\mathrm{i}}{2} \Pi_y^{-1} V_{ij} p \end{pmatrix}\\ &= \begin{pmatrix} \Pi_y^{-1} V_{ij} p \\ -\Pi_x \Pi_y^{-1} V_{ij} p + \mathrm{i}V_{ij} p \end{pmatrix}. \end{align}\] The symplectic connection is given by \[\partial_{ij}\mapsto \partial_{ij}+\mathcal{V},\text{ where } \mathcal{V} = \Pi_y^{-1} V_{ij} p \cdot \partial_\xi + (-\Pi_x \Pi_y^{-1} V_{ij} p + \mathrm{i}V_{ij} p)\cdot \partial_\eta.\] Recall that \[q= \tfrac{1}{2}(\eta + \bar{\eta}) + \tfrac{1}{2}\Pi(\xi + \bar{\xi}),\quad p=-\tfrac{1}{2} \Pi_y^{-1} \bigl( (\eta - \bar{\eta}) + \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Pi} \endgroup (\xi - \bar{\xi}) \bigr).\] Then we have \[\begin{align} \mathcal{V}(q) &= \tfrac{1}{2} \mathcal{V}(\eta) + \tfrac{1}{2} \Pi \mathcal{V}(\xi) = \tfrac{1}{2} \Gamma_{ij}^\eta + \tfrac{1}{2} \Pi \Gamma_{ij}^\xi \\ &= \tfrac{1}{2} \bigl( (-\Pi_x \Pi_y^{-1} V_{ij} p + \mathrm{i}V_{ij} p) + (\Pi_x + \mathrm{i}\Pi_y)(\Pi_y^{-1} V_{ij} p) \bigr) \\ &= \tfrac{1}{2} \bigl( -\Pi_x \Pi_y^{-1} V_{ij} p + \mathrm{i}V_{ij} p + \Pi_x \Pi_y^{-1} V_{ij} p + \mathrm{i}\Pi_y \Pi_y^{-1} V_{ij} p \bigr) \\ &= \tfrac{1}{2} ( \mathrm{i}V_{ij} p + \mathrm{i}V_{ij} p ) = \mathrm{i}V_{ij} p. \end{align}\] Similarly we have \[\begin{align} \mathcal{V}(\bar{q}) &= \tfrac{1}{2} \mathcal{V}(\eta) + \tfrac{1}{2} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Pi} \endgroup \mathcal{V}(\xi) = \tfrac{1}{2} \Gamma_{ij}^\eta + \tfrac{1}{2} \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Pi} \endgroup \Gamma_{ij}^\xi \\ &= \tfrac{1}{2} \bigl( (-\Pi_x \Pi_y^{-1} V_{ij} p + \mathrm{i}V_{ij} p) + (\Pi_x - \mathrm{i}\Pi_y)(\Pi_y^{-1} V_{ij} p) \bigr) \\ &= \tfrac{1}{2} \bigl( -\Pi_x \Pi_y^{-1} V_{ij} p + \mathrm{i}V_{ij} p + \Pi_x \Pi_y^{-1} V_{ij} p - \mathrm{i}V_{ij} p \bigr) \\ &= 0. \end{align}\] Now applying the above two equalities \(\mathcal{V}(q)=\mathrm{i}V_{ij}p\) and \(\mathcal{V}(\bar{q})=0\), we have \[\begin{align} \mathcal{V}(p) &= -\tfrac{1}{2} \Pi_y^{-1} \bigl( \mathcal{V}(\eta) + \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Pi} \endgroup \mathcal{V}(\xi) \bigr) = -\Pi_y^{-1} \mathcal{V}(\bar{q})=0,\\ \mathcal{V}(\bar{p}) &= \tfrac{1}{2} \Pi_y^{-1} \bigl( \mathcal{V}(\eta) + \Pi \mathcal{V}(\xi) \bigr) = \Pi_y^{-1} \mathcal{V}(q)=\mathrm{i}\Pi_y^{-1} V_{ij} p. \end{align}\] Combining these together, we obtain \[\mathcal{V}=\mathcal{V}(q)\cdot\partial_q+\mathcal{V}(\bar{p})\cdot\partial_{\bar{p}}=\mathrm{i}V_{ij} p\cdot \partial_q +\mathrm{i}\Pi_y^{-1} V_{ij} p\cdot \partial_{\bar{p}}.\]

To show \(\partial_A-\partial_{\omega_{X/S}^A}=2\sqrt{2}\beta_{X/S}^{-1}(\theta)\), it remains to check that \[2(\id-J_\Pi^C)\bigl(-\tfrac{\mathrm{i}}{2} V_{ij} p \cdot \partial_q\bigr)=-(\mathrm{i}V_{ij} p\cdot \partial_q +\mathrm{i}\Pi_y^{-1} V_{ij} p\cdot \partial_{\bar{p}}).\] This follows since \[\mathrm{i}J_\Pi^C(V_{ij}p\cdot \partial_q)=\mathrm{i}J_\Pi^C(V_{ij}p\cdot (\partial_{q_x}-\mathrm{i}\partial_{q_y}))=-\mathrm{i}\Pi_y^{-1}V_{ij}p (\partial_{p_x}+\mathrm{i}\Pi_y^{-1}\partial_{p_y})=-\mathrm{i}\Pi_y^{-1} V_{ij} p\cdot \partial_{\bar{p}}.\] We prove the last statement. Since \(\theta(\partial_{ij})\) is \(J_\Pi^B\)-holomorphic, we have \(J_\Pi^B (\theta(\partial_{ij})) = \mathrm{i}(\theta(\partial_{ij}))\) and \[J_\Pi^A (\theta(\partial_{ij}))=-J_\Pi^C (\mathrm{i}\theta(\partial_{ij}))=\tfrac{1}{2}\Pi_y^{-1} V_{ij} p\cdot \partial_{\bar{p}}.\] Then we have \[2(a_0\id+a_1J_\Pi^A+a_2J_\Pi^B+a_3J_\Pi^C)(\theta(\partial_{ij}))=(a_2-\mathrm{i}a_0)V_{ij}p\cdot\partial_q+(a_1+\mathrm{i}a_3)\Pi_y^{-1}V_{ij}p\cdot\partial_{\bar{p}}.\] By comparing the coefficients, we see that \((\beta_{X/S}^\RR)^{-1}\) must be the one in ?? . ◻

Proof. We define the gauge transformation \(\Phi := \beta_{X/S,\mathrm{can}}^{\RR} \comp (\beta_{X/S}^\RR)^{-1}\), which is clearly a unitary automorphism of \((T_{X/S}^\RR,g_{X/S},J_{X/S}^B)\). The vectors \(\{\theta(\partial_{ij})_\RR\}\) span \(E_q\) (at generic points where \(p \ne 0\)), so the condition \(\beta_{X/S}^{-1}(\theta) = (\beta_{X/S,\mathrm{can}})^{-1}(\theta)\) implies \(\Phi v = v\) for all \(v\in E_{\mathrm{ab}}\). Since the splitting \(T_{X/S}^\RR=E_{\mathrm{ab}}\oplus E_{\mathrm{fib}}\) is orthogonal with respect to \(g_{X/S}\) and is preserved by \(J_{X/S}^B\), \(\Phi\) has the block diagonal form as above. ◻

Proof. In adapted local holomorphic coordinates, we may write \[\nabla_2^\RR(\partial_{x^i})= \partial_{x^i}+\Gamma_{x^i}^{u^\alpha}\partial_{u^\alpha}+\Gamma_{x^i}^{w^\alpha}\partial_{w^\alpha},\quad \nabla_2^\RR(\partial_{y^i})= \partial_{y^i}+\Gamma_{y^i}^{u^\alpha}\partial_{u^\alpha}+\Gamma_{y^i}^{w^\alpha}\partial_{w^\alpha},\] where \(s^i=x^i+\mathrm{i}y^i\), \(z^\alpha=u^\alpha+\mathrm{i}w^\alpha\). Then we have \[\begin{align} \nabla_2^{1,0}(\partial_i)&=\partial_i+\tfrac{1}{2}(\Gamma_{x^i}^{u^\alpha}-\mathrm{i}\Gamma_{y^i}^{u^\alpha})(\partial_\alpha+\partial_{\bar{\alpha}})+\tfrac{1}{2}(\Gamma_{x^i}^{w^\alpha}-\mathrm{i}\Gamma_{y^i}^{w^\alpha})\mathrm{i}(\partial_\alpha-\partial_{\bar{\alpha}})\\ &=\partial_i+\tfrac{1}{2}(\Gamma_{x^i}^{u^\alpha}-\mathrm{i}\Gamma_{y^i}^{u^\alpha}+\mathrm{i}\Gamma_{x^i}^{w^\alpha}+\Gamma_{y^i}^{w^\alpha})\partial_{\alpha}+\tfrac{1}{2}(\Gamma_{x^i}^{u^\alpha}-\mathrm{i}\Gamma_{y^i}^{u^\alpha}-\mathrm{i}\Gamma_{x^i}^{w^\alpha}-\Gamma_{y^i}^{w^\alpha})\partial_{\bar{\alpha}}. \end{align}\] By the condition \(\nabla_1^\RR(v)=\frac{1}{2}(\nabla_2^\RR(v)-J\nabla_2^\RR(J_S v))\), we have \[\begin{align} \nabla_1^{\RR}(\partial_{x^i})&=\tfrac{1}{2}(\partial_{x^i}+\Gamma_{x^i}^{u^\alpha}\partial_{u^\alpha}+\Gamma_{x^i}^{w^\alpha}\partial_{w^\alpha})-\tfrac{1}{2}J(\partial_{y^i}+\Gamma_{y^i}^{u^\alpha}\partial_{u^\alpha}+\Gamma_{y^i}^{w^\alpha}\partial_{w^\alpha})\\ &=\tfrac{1}{2}(\partial_{x^i}+\Gamma_{x^i}^{u^\alpha}\partial_{u^\alpha}+\Gamma_{x^i}^{w^\alpha}\partial_{w^\alpha})-\tfrac{1}{2}(-\partial_{x^i}+\Gamma_{y^i}^{u^\alpha}\partial_{w^\alpha}-\Gamma_{y^i}^{w^\alpha}\partial_{u^\alpha})\\ &=\partial_{x^i}+\tfrac{1}{2}(\Gamma_{x^i}^{u^\alpha}+\Gamma_{y^i}^{w^\alpha})\partial_{u^\alpha}+\tfrac{1}{2}(\Gamma_{x^i}^{w^\alpha}-\Gamma_{y^i}^{u^\alpha})\partial_{w^\alpha}. \end{align}\] Similarly, \[\nabla_1^{\RR}(\partial_{y^i})=\partial_{y^i}+\tfrac{1}{2}(\Gamma_{y^i}^{u^\alpha}-\Gamma_{x^i}^{w^\alpha})\partial_{u^\alpha}+\tfrac{1}{2}(\Gamma_{x^i}^{u^\alpha}+\Gamma_{y^i}^{w^\alpha})\partial_{w^\alpha}.\] Then we have \[\nabla_1^{1,0}(\partial_i)=\partial_i+\tfrac{1}{2}(\Gamma_{x^i}^{u^\alpha}+\Gamma_{y^i}^{w^\alpha}+\mathrm{i}(\Gamma_{x^i}^{w^\alpha}-\Gamma_{y^i}^{u^\alpha}))\partial_{\alpha}.\] By taking the conjugate, we see that \(\nabla_1^{0,1}\) induces \(\bar{\partial}_f\). By comparing the above formulas for \(\nabla_1^{1,0}\) and \(\nabla_2^{1,0}\), we see that they induce the same almost connection.

For the last statement, we have \[(\bar{\partial}_2-\bar{\partial}_f)(\partial_{\bar{i}})= \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\nabla_2^{1,0}(\partial_i)-\nabla_1^{1,0}(\partial_i)} \endgroup =\tfrac{1}{2}(\Gamma_{x^i}^{u^\alpha}+\mathrm{i}\Gamma_{y^i}^{u^\alpha}+\mathrm{i}\Gamma_{x^i}^{w^\alpha}-\Gamma_{y^i}^{w^\alpha})\partial_{\alpha}.\] On the other hand, \[\begin{align} \tfrac{1}{2}(\nabla_2^\RR(\partial_{x^i})+J\nabla_2^\RR(\partial_{y^i}))^{1,0}&=\tfrac{1}{2}(\Gamma_{x^i}^{u^\alpha}\partial_{u^\alpha}+\Gamma_{x^i}^{w^\alpha}\partial_{w^\alpha}+\Gamma_{y^i}^{u^\alpha}\partial_{w^\alpha}-\Gamma_{y^i}^{w^\alpha}\partial_{u^\alpha})^{1,0}\\ &=\tfrac{1}{2}(\Gamma_{x^i}^{u^\alpha}+\mathrm{i}\Gamma_{y^i}^{u^\alpha}+\mathrm{i}\Gamma_{x^i}^{w^\alpha}-\Gamma_{y^i}^{w^\alpha})\partial_{\alpha}. \end{align}\] Therefore 80 is proved. ◻

Proof. We compute \[J_A\bigl( \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{w_B^{1,0}} \endgroup \bigr) = \tfrac{1}{2}J_A (w + \mathrm{i}J_B w) = \tfrac{1}{2} (J_A w + \mathrm{i}J_C w).\] Now the equation \(v_B^{1,0} = J_A\bigl( \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{w_B^{1,0}} \endgroup \bigr)\) is equivalent to \[\tfrac{1}{2}(v - \mathrm{i}J_B v) = \tfrac{1}{2} (J_A w + \mathrm{i}J_C w).\] Equating the real and imaginary parts, we obtain \(v = J_A w\) and \(-J_B v = J_C w\). It remains to check the first equation implies the second. This is true since \(-J_B v = -J_B (J_A w) = J_C w\). ◻

Proof. By 80 and Lemma 50, we only need to compute \(\frac{1}{2}(\ell_{[\mu]}+\mathrm{I}\ell_{[\mathrm{i}\mu]})\). Let \((\mu, \beta_1, \psi_1)\) be the semiharmonic representative of \(\ell_{[\mu]}\) and \((\mathrm{i}\mu, \beta_2, \psi_2)\) be that of \(\ell_{[\mathrm{i}\mu]}\). Then \(\mathrm{I} \ell_{[\mathrm{i}\mu]}\) is represented by \((-\mu, \mathrm{i}\beta_2, \mathrm{i}\psi_2)\), and \(\frac{1}{2}(\ell_{[\mu]}+\mathrm{I}\ell_{[\mathrm{i}\mu]})\) is represented by \(\frac{1}{2}(0,\beta_1+\mathrm{i}\beta_2,\psi_1+\mathrm{i}\psi_2)\). We have \[\tfrac{1}{2}J_{X/S}^A(\ell_{[\mu]}+\mathrm{I}\ell_{[\mathrm{i}\mu]})=\tfrac{1}{2}[(\mathrm{i}\psi_1^*+\psi_2^*,-\mathrm{i}\beta_1^*-\beta_2^*)].\] By [18], there exist unique Hermitian sections \(\zeta_1, \zeta_2\in C^\infty(\Sigma,\ad P)\) such that \[D''\zeta_1 = \tfrac{1}{2\mathrm{i}} \bigl(\beta_1^* - \psi_1^* + \tfrac{1}{2\mathrm{i}}\Phi\mu\bigr)\quad\text{and}\quad D''\zeta_2 = \tfrac{1}{2\mathrm{i}} \bigl(\beta_2^* - \psi_2^* + \tfrac{1}{2\mathrm{i}}\Phi\mathrm{i}\mu\bigr).\] Therefore, we have \[D''(\zeta_1-\mathrm{i}\zeta_2)=\tfrac{1}{2}(\mathrm{i}\psi_1^*+\psi_2^*-\mathrm{i}\beta_1^*-\beta_2^*)-\mathrm{i}\tfrac{1}{2\mathrm{i}}\Phi\mu.\] This implies \[\tfrac{1}{2}[(\mathrm{i}\psi_1^*+\psi_2^*,-\mathrm{i}\beta_1^*-\beta_2^*)]=\bigl[\bigl(\tfrac{\mathrm{i}}{2\mathrm{i}}\Phi\mu,0\bigr)\bigr]=\mathrm{I}\Theta([\mu]).\] By 80 and Lemma 50, we have \[(\bar{\partial}_{B,0}-\bar{\partial}_B)([\mu]^{0,1})=\tfrac{1}{2}(\ell_{[\mu]}+\mathrm{I}\ell_{[\mathrm{i}\mu]})_B^{1,0}=J_{X/S}^A( \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{(-{\mathrm{I}}\Theta([\mu]))_B^{1,0}} \endgroup )=\mathrm{i}J_{X/S}^A( \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta([\mu])_B^{1,0}} \endgroup ).\] Therefore, \(\bar{\partial}_{B,0}-\bar{\partial}_B= \begingroup \def\mathaccent##1##2{ \kern 0.8\dimexpr\macc@kerna \overline{\kern-0.8\dimexpr\macc@kerna\macc@nucleus\kern 0.2\dimexpr\macc@kerna} \kern-0.2\dimexpr\macc@kerna } \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax} \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a \macc@nested@a\relax 111{\Theta} \endgroup _{J}\). ◻

Proof. Let \(f = \phi + \mathrm{i}\psi\), where \(\phi = \Re(f)\) and \(\psi = \Im(f)\). The condition \(\omega_{J_A}(u, v) = \mathrm{i}\bar{\partial}_{J_A} f (v)\) implies \(g_Y(J_A u, v) = \mathrm{i}v(f)=\mathrm{i}g_Y(u, v)\). Note that \(v=\frac{1}{2}(v_\RR+\mathrm{i}J_Av_\RR)\). Taking the real part of \(g_Y(u, v) = v(f)\) yields \[\begin{gather} g_Y(u_\RR, v_\RR) = v_\RR(\phi) - (J_A v_\RR)(\psi)\\=g_Y(\grad \phi,v_\RR)-g_Y(\grad\psi,J_A v_\RR)=g_Y(\grad \phi + J_A \grad \psi, v_\RR). \end{gather}\] Since \(v_\RR\) is arbitrary, \(u_\RR=\grad \phi + J_A \grad \psi\).

The equation \(u = 2(\id - J_C)(w)\) is equivalent to \[\begin{align} (u_\RR - \mathrm{i}J_A u_\RR) &= 2(\id - J_C)(w_\RR - \mathrm{i}J_B w_\RR) \\ &= 2w_\RR - 2J_C w_\RR - 2\mathrm{i}(J_B w_\RR +J_A w_\RR). \end{align}\] Comparing the real and imaginary parts, we obtain \(u_\RR = 2(\id - J_C) w_\RR\) and \(J_Au_\RR = 2 (J_A+J_B) w_\RR\). The first equation implies the second, and is equivalent to \[\begin{align} w_\RR &= \tfrac{1}{2} (\id - J_C)^{-1} u_\RR = \tfrac{1}{4} (\id + J_C) u_\RR\\ &=\tfrac{1}{4} (\id + J_C)(\grad \phi + J_A \grad \psi) \\ &=\tfrac{1}{4} \left( \grad \Re(f) + J_C \grad \Re(f) + J_A \grad \Im(f) + J_B \grad \Im(f) \right). \qedhere \end{align}\] ◻

Proof. Denote \(\mathcal{V}:=(\partial_A-\partial_{\omega_{X/S}^A})([\mu]^{1,0})\). By the orthogonality condition satisfied by the symplectic connection, we have \(\omega(\nabla_{\omega_{X/S}^A}^{1,0}([\mu^{1,0}]),v)=0\) for any vertical \(J_{X/S}^{A}\)-\((1,0)\)-vector field \(v\). This implies \[\omega_{X/S}^A(\mathcal{V},v)=\omega(\nabla^{1,0}([\mu]^{1,0}),v)=\mathrm{i}\bar{\partial}_{J_{X/S}^A}\bigl(\tfrac{1}{2}(\mathrm{d}\mathrm{E}(\ell_{[\mu]})-\mathrm{i}\mathrm{d}\mathrm{E}(\ell_{[\mathrm{i}\mu]}))\bigr)(v),\] where the second equality follows from the definition \(\omega=\mathrm{i}\partial_{X_A}\bar{\partial}_{X_A}\mathrm{E}\). By Lemma 52, it suffices to prove \[\Theta([\mu])=\tfrac{1}{8}(\grad \mathrm{d}\mathrm{E}(\ell_{[\mu]}) + J_{X/S}^C \grad \mathrm{d}\mathrm{E}(\ell_{[\mu]})-J_{X/S}^A\grad \mathrm{d}\mathrm{E}(\ell_{[\mathrm{i}\mu]})-J_{X/S}^B\grad \mathrm{d}\mathrm{E}(\ell_{[\mathrm{i}\mu]})).\] By Lemma 53, \(\mathrm{d}\mathrm{E}(\ell_{[\mu]})=2\Re\bigl\langle \Phi, \bigl(\tfrac{1}{2\mathrm{i}}\Phi\mu\bigr)^*\bigr\rangle\). Let \(w=(\beta,\psi)\) be a harmonic vertical tangent vector. Then \[\mathrm{d}(\mathrm{d}\mathrm{E}(\ell_{[\mu]}))(w) = 2\Re\bigl( \bigl\langle \psi, \bigl(\tfrac{1}{2\mathrm{i}}\Phi\mu\bigr)^* \bigr\rangle + \bigl\langle \Phi, \bigl(\tfrac{1}{2\mathrm{i}}\psi\mu\bigr)^* \bigr\rangle \bigr) = 4\Re\bigl\langle \psi, \bigl(\tfrac{1}{2\mathrm{i}}\Phi\mu\bigr)^* \bigr\rangle,\] and \(\grad \mathrm{d}\mathrm{E}(\ell_{[\mu]})=4[(0,(\frac{1}{2\mathrm{i}}\Phi\mu)^*)]\). Replacing \(\mu\) by \(\mathrm{i}\mu\), we have \(\grad \mathrm{d}\mathrm{E}(\ell_{[\mathrm{i}\mu]})=4[(0,(\frac{\mathrm{i}}{2\mathrm{i}}\Phi\mu)^*)]\). Consequently, \[\begin{gather} \tfrac{1}{8}(\grad \mathrm{d}\mathrm{E}(\ell_{[\mu]}) + J_{X/S}^C \grad \mathrm{d}\mathrm{E}(\ell_{[\mu]})-J_{X/S}^A\grad \mathrm{d}\mathrm{E}(\ell_{[\mathrm{i}\mu]})-J_{X/S}^B\grad \mathrm{d}\mathrm{E}(\ell_{[\mathrm{i}\mu]}))\\ =\tfrac{1}{2}\bigl[\bigl(0,\bigl(\tfrac{1}{2\mathrm{i}}\Phi\mu\bigr)^*\bigr)+\bigl(\tfrac{1}{2\mathrm{i}}\Phi\mu,0\bigr)+\bigl(\tfrac{1}{2\mathrm{i}}\Phi\mu,0\bigr)-\bigl(0,\bigl(\tfrac{1}{2\mathrm{i}}\Phi\mu\bigr)^*\bigr)\bigr]=\bigl[\bigl(\tfrac{1}{2\mathrm{i}}\Phi\mu,0\bigr)\bigr]=\Theta([\mu]). \end{gather}\] The proposition is proved. ◻

Proof. This is a local problem, so we may assume that \(f_{\mathrm{dR}}\) is a trivial bundle \(S\times M_{\mathrm{dR}}(X_{s_0},G)\to S\). Consider the map \[F:M_{\mathrm{Dol}}(X/S,G)\to S\times M_{\mathrm{dR}}(X_{s_0},G),\quad x\mapsto (f_{\mathrm{Dol}}(x),\mathrm{H}(x)),\] where \(\mathrm{H}\) is the nonabelian Hodge map. \(F\) is a homeomorphism. By [18], \(F\) is real analytic. If \(\mathrm{d}F_x(v)=(0,0)\), then \(\mathrm{d}f_{\mathrm{Dol},x}(v)=0\), which means \(v\) is a vertical tangent vector. Since \(\mathrm{H}\) is a real analytic diffeomorphism, we have \(v=0\). Therefore, \(\mathrm{d}F\) is invertible. By the real analytic inverse function theorem, \(F^{-1}\) is also real analytic. Hence, \(f_{\mathrm{dR}}\) and \(f_{\mathrm{Dol}}\) are isomorphic smooth fiber bundles. ◻

Proof. The problem is local. For a given point \(s\in S\), take an open neighborhood \(s\in U\subset S\) so that the family \(X\to S\) over \(U\) admits a smooth trivialization \(f^{-1}(U)\cong U\times \Sigma\) over \(U\). Then in the semisimple case, there is a unique holomorphic map \(\phi: U\to \mathbf{T}(\Sigma)\) such that the pullback family \(M(G)\times_{\mathbf{T}(\Sigma)}U \to U\) is isomorphic to \(f^{-1}(U)\to U\). Note that the (twisted) Simpson mechanism is compatible with pullback. The result follows from the universal case as carried out in Section 6.3.

For the rank one case, i.e., \(G=\CC^*\), we consider the associated weight one \(\ZZ\)-PVHS \(V_{\ZZ}\) to \(X\to S\). The theory of variation of Hodge structure applied to \(V_{\ZZ}\) gives rise to a linear flat bundle \((V,\nabla_{\mathrm{GM}})\) and a linear Higgs bundle \((E=E^{1,0}\oplus E^{0,1},\theta_{\mathrm{KS}})\). They correspond to each other under the nonabelian Hodge correspondence. This holds true for a general weight one \(\ZZ\)-PVHS, namely a polarized family of abelian varieties \(A\to S\) (In our case, it is a polarized family of Jacobians). As \(V_{\ZZ}\subset V\) and \(\nabla_{\mathrm{GM}}\) is linear, we obtain a nonlinear flat bundle by taking the quotient \((V/V_{\ZZ}\to S, \overline{\nabla}_{\mathrm{GM}})\), which is nothing but the rank one relative de Rham moduli equipped with the nonabelian Gauss-Manin connection. We know that the relative Dolbeault moduli is the cotangent bundle of the relative Picard variety \(\textrm{Pic}^0(X/S)\). Analytically, it is isomorphic to the quotient \(E^{1,0}\times E^{0,1}/V_{\ZZ}\to S\). In general, it is the relative cotangent bundle \(T^*_{A/S}\to S\). By linearity, the linear Higgs field \(\theta_{\mathrm{KS}}\) descends to a nonlinear Higgs field on the relative cotangent bundle, which in the case of rank one relative Dolbeault moduli is nothing but the nonabelian Kodaira-Spencer map. Now the result follows from the universal one verified in Section 6.2. ◻