By assumption there is a 1-form $\zeta$ on $U$ such that

$$\omega_{SF}-\omega=d\zeta=\partial\zeta^{0,1}+\overline{\partial}\zeta^{1,0},\ \ \ \overline{\partial}\zeta^{0,1}=0,$$

where $\zeta=\zeta^{0,1}+\zeta^{1,0}$ and $\zeta^{0,1}=\overline{\zeta^{1,0}}$.

We claim that $(0,1)$-forms

(3.4)

$$\theta_{j}=\sqrt{-1}\ \overline{\partial}\left(\sum_{i=1}^{n-m}g_{ij}(y)(z_{i}-\bar{z}_{i})\right),\ \ \ \ \ \ j=1,\cdots,n-m,$$

are invariant under translations by flat sections of the Gauss-Manin connection on $B\times\mathbb{C}^{n-m}$, and thus descend to $(0,1)$-forms on $U$. It is enough to check invariance under translation by $\lambda s$ where $s$ is a generator of $\Lambda$ and $\lambda\in\mathbb{R}$. First, consider a general translation $z_{i}\mapsto z_{i}+\tau_{i}(y)$. If $\tau_{i}=\lambda\delta_{ik}$ for some $k$, so that $\tau_{i}$ is real, then $\theta_{j}$ are invariant. If $\tau_{i}=\lambda Z_{ik}$ for some $k$, $\lambda\in\mathbb{R}$, we obtain

$$\begin{split}\sum_{i=1}^{n-m}g_{ij}(z_{i}+\lambda Z_{ik}-\bar{z}_{i}-\lambda\bar{Z}_{ik})&=\sum_{i=1}^{n-m}g_{ij}(z_{i}-\bar{z}_{i})\\ &\ \ \ +2\sqrt{-1}\sum_{i=1}^{n-m}({\rm Im}Z)^{-1}_{ij}\lambda({\rm Im}Z)_{ik}\\ &=\sum_{i=1}^{n-m}g_{ij}(z_{i}-\bar{z}_{i})+2\lambda\sqrt{-1}\delta_{jk}.\end{split}$$

Applying $\bar{\partial}$ kills the correction term, so $\theta_{j}$ are invariant, and therefore they define $(0,1)$-forms on $U$. Since, for any $y\in B$,

(3.5)

$$p^{*}\left(\theta_{j}|_{M_{y}}\right)=-\sqrt{-1}\sum_{i=1}^{n-m}g_{ij}(y)d\bar{z}_{i},$$

is fiberwise constant and $g_{ij}$ is non-degenerate, we have that $[\theta_{i}|_{M_{y}}]$, $i=1,\cdots,n-m$ is a basis of $H^{0,1}(M_{y})$.

We claim that there are holomorphic functions $\sigma_{i}:B\rightarrow\mathbb{C}$ such that

(3.6)

$$\zeta^{0,1}=\sum_{i=1}^{n-m}\sigma_{i}\theta_{i}+\overline{\partial}h,$$

for a complex-valued function $h$ on $U$. To prove this, note that $H^{0,1}(U)=H^{1}(U,\mathcal{O}_{U})$ which by the Leray spectral sequence for $f$ is isomorphic to $H^{0}(B,R^{1}f_{*}\mathcal{O}_{U})$ since $H^{k}(B,f_{*}\mathcal{O}_{U})=H^{k}(B,\mathcal{O}_{B})=0$ for $k\geqslant 1$. It follows that a $\overline{\partial}$-closed $(0,1)$-form on $U$ represents the zero class if and only if its restriction to $M_{y}$ represents the zero class in $H^{0,1}(M_{y})$ for all $y\in B$. Consider now the $(0,1)$-forms $d\overline{y}^{i}$, $1\leqslant i\leqslant m$, on $B$ and denote their pullbacks to $U$ by the same symbol. Then at each point of $U$ the forms $\{\theta_{j}\},1\leqslant j\leqslant n-m$ together with $\{d\overline{y}^{i}\},1\leqslant i\leqslant m$, form a basis of $(0,1)$-forms. We can then write

$$\zeta^{0,1}=\sum_{j=1}^{n-m}w_{j}\theta_{j}+\sum_{i=1}^{m}h_{i}d\overline{y}^{i},$$

where $w_{j},h_{i}$ are smooth complex functions on $U$. If we now restrict to a fiber $M_{y}$ we get $\zeta^{0,1}|_{M_{y}}=\sum_{j=1}^{n-m}w_{j}\theta_{j}|_{M_{y}},$ and the functions $w_{j}$ restricted to $M_{y}$ can be thought of as functions on $\mathbb{C}^{n-m}$ which are periodic with period $\Lambda_{y}$. There is a holomorphic $T^{2n-2m}$-action on $U$ which is induced by the action of $\mathbb{R}^{2n-2m}$ on $B\times\mathbb{C}^{n-m}$ given by $x\cdot(y,z)=(y,z+\sum_{j}x_{j}\tau_{j}(y))$, where $\tau_{j}(y)$ is a basis for the lattice $\Lambda_{y}$ (the choice of which is irrelevant). If $\alpha$ is a function or differential form on $U$ or $M_{y}$, we will denote by $\tilde{\alpha}$ its average with respect to the $T^{2n-2m}$-action. In particular, if $\alpha$ is a function on $U$ then $\tilde{\alpha}$ is the pullback of a function from $B$. We now call $\sigma_{j}=\tilde{w}_{j}$, $1\leqslant j\leqslant n-m$, which are functions of $y\in B$ only. We clearly have that $\tilde{\theta}_{j}=\theta_{j}$ and $\widetilde{d\overline{y}^{i}}=d\overline{y}^{i}$, so

$$\widetilde{\zeta^{0,1}|_{M_{y}}}=\sum_{j=1}^{n-m}\sigma_{j}(y)\theta_{j}|_{M_{y}}.$$

Now the $T^{2n-2m}$-action on $M_{y}$ is generated by holomorphic vector fields and therefore acts trivially on the Dolbeault cohomology $H^{0,1}(M_{y})$, which implies that

$$\left[\zeta^{0,1}|_{M_{y}}\right]=\left[\widetilde{\zeta^{0,1}|_{M_{y}}}\right]=\sum_{j=1}^{n-m}\sigma_{j}(y)\left[\theta_{j}|_{M_{y}}\right],$$

in $H^{0,1}(M_{y})$ for all $y\in B$. If we show that the $\sigma_{j}(y)$ are holomorphic, then the $(0,1)$-form $\zeta^{0,1}-\sum_{j}\sigma_{j}(y)\theta_{j}$ on $U$ would be $\overline{\partial}$-closed and cohomologous to zero in $H^{0,1}(U)$, thus proving (3.6).

Call now $V_{j}$, $1\leqslant j\leqslant n-m$ and $W_{i}$, $1\leqslant i\leqslant m$ the $T^{2n-2m}$-invariant $(0,1)$-type vector fields on $U$ which are the dual basis to $\theta_{j},d\overline{y}^{i}$. We have that $V_{j}=\sqrt{-1}\sum_{k=1}^{n-m}g^{jk}\frac{\partial}{\partial\overline{z}_{k}}$, where $g^{jk}$ is the inverse matrix of $g_{jk}$, and the vector fields $\frac{\partial}{\partial\overline{z}_{k}}$ are well-defined on $U$. We will not need the explicit formula for $W_{i}$, but just the fact that if a function $f$ on $U$ is the pullback of a function on $B$ then $W_{i}(f)=\frac{\partial f}{\partial\overline{y}_{i}}$.

To see why $\sigma_{j}(y)$ is holomorphic, compute

$$\begin{split}0=\overline{\partial}\zeta^{0,1}=&\sum_{i,j}W_{i}(w_{j})d\overline{y}^{i}\wedge\theta_{j}+\sum_{i,j}V_{i}(w_{j})\theta_{i}\wedge\theta_{j}\\ &+\sum_{i,j}W_{j}(h_{i})d\overline{y}^{j}\wedge d\overline{y}^{i}+\sum_{i,j}V_{j}(h_{i})\theta_{j}\wedge d\overline{y}^{i}.\end{split}$$

Since each $V_{j}$ is a linear combination of $\frac{\partial}{\partial\overline{z}_{k}}$, we have that the functions $V_{i}(w_{j})$ and $V_{j}(h_{i})$ have average zero on each fiber. Taking the average then gives

$$0=\overline{\partial}\widetilde{\zeta^{0,1}}=\sum_{i,j}\frac{\partial\sigma_{j}}{\partial\overline{y}_{i}}d\overline{y}^{i}\wedge\theta_{j}+\sum_{i,j}\frac{\partial\tilde{h}_{i}}{\partial\overline{y}_{j}}d\overline{y}^{j}\wedge d\overline{y}^{i}.$$

Since the forms $d\overline{y}^{i}\wedge\theta_{j}$ and $d\overline{y}^{j}\wedge d\overline{y}^{i}$ are linearly independent at every point, this implies that $\sigma_{j}(y)$ are indeed holomorphic.

Let now $T_{\sigma}:U\to U$ be the translation induced by the section $\sigma=(p\circ\sigma_{1},\cdots,p\circ\sigma_{n-m})$, where $p:B\times\mathbb{C}^{n-m}\rightarrow U$ is the quotient map. Since

$$\begin{split}\sum_{i,j}&-{\frac{g_{ij}}{2}}\left((z_{i}+\sigma_{i}-\bar{z}_{i}-\bar{\sigma}_{i})(z_{j}+\sigma_{j}-\bar{z}_{j}-\bar{\sigma}_{j})\right)\\ ={}&\eta-\sum_{i,j}{\frac{g_{ij}}{2}}\left((\sigma_{i}-\bar{\sigma}_{i})(z_{j}-\bar{z}_{j})+(\sigma_{j}-\bar{\sigma}_{j})(z_{i}-\bar{z}_{i})+(\sigma_{i}-\bar{\sigma}_{i})(\sigma_{j}-\bar{\sigma}_{j})\right)\\ ={}&\eta-\sum_{i,j}g_{ij}\left((\sigma_{i}-\bar{\sigma}_{i})(z_{j}-\bar{z}_{j})+{\frac{1}{2}}(\sigma_{i}-\bar{\sigma}_{i})(\sigma_{j}-\bar{\sigma}_{j})\right),\end{split}$$

we have

$$\begin{split}p^{*}T_{\sigma}^{*}\omega_{SF}-p^{*}\omega_{SF}&=-\sqrt{-1}\partial\overline{\partial}\sum_{i,j}g_{ij}(\sigma_{i}-\bar{\sigma}_{i})(z_{j}-\bar{z}_{j})+\sqrt{-1}\partial\overline{\partial}\Phi(y)\\ &=p^{*}\left(-\partial\sum_{i}\sigma_{i}\theta_{i}-\overline{\partial}\sum_{i}\overline{\sigma_{i}\theta_{i}}\right)+\sqrt{-1}\partial\overline{\partial}\Phi(y),\end{split}$$

where $\Phi(y)=-\sum_{i,j}\frac{g_{ij}}{2}(\sigma_{i}-\bar{\sigma}_{i})(\sigma_{j}-\bar{\sigma}_{j})$ is a real function of $y$ only. We have just proved that

$$\omega_{SF}-\omega=\partial\zeta^{0,1}+\overline{\partial}\ \overline{\zeta^{0,1}}=\partial\sum_{i}\sigma_{i}\theta_{i}+\partial\overline{\partial}h+\overline{\partial}\sum_{i}\overline{\sigma_{i}\theta_{i}}+\overline{\partial}\partial\overline{h}.$$

Thus

$$p^{*}T_{\sigma}^{*}\omega_{SF}-p^{*}\omega=p^{*}\sqrt{-1}\partial\overline{\partial}(2\mathrm{Im}h+\Phi),$$

which proves (3.3) with $\xi=2\mathrm{Im}h+\Phi$. ∎