we will encounter in the proof of Theorem \[thm:kaehlerity\]. For now we observe that the vanishing of these sections indeed imply that $(M, g(\tau))$ will remain Kähler relative to the fixed complex structure $J(0) = J_0$.

\[lem:jext\] Let $g$, $J$, ${\hat{{\operatorname{R}}}}$, and ${\hat{{\operatorname{P}}}}$ be as above. If ${\hat{{\operatorname{R}}}}= 0$ and $\nabla {\hat{{\operatorname{P}}}}= 0$ on $M\times [0, T]$, and $\nabla J = 0$ on $M\times\{0\}$, then $\nabla J = 0$ and ${{\frac{\partial }{\partial\tau}}}J = 0$ on $M\times[0, T]$.

First observe that our assumptions imply that ${\operatorname{S}}\circ {\operatorname{F}}= 0$. Indeed, $${\operatorname{S}}\circ {\operatorname{F}}= {\operatorname{S}}\circ ({\hat{{\operatorname{P}}}}\circ {\operatorname{F}}) = {\hat{{\operatorname{S}}}}\circ {\operatorname{F}}= (\nabla {\operatorname{R}}\circ {\hat{{\operatorname{P}}}}) \circ {\operatorname{F}}= \nabla {\hat{{\operatorname{R}}}}\circ {\operatorname{F}}= 0,$$ using Lemma \[lem:projprop\]. The observation that $J$ remains parallel then follows from a direct pointwise calculation. Fix $p\in M$ and let $W_{ib}^a = \nabla_iJ^b_a$. Since $D_{\tau} J = 0$, we have $$\begin{aligned}
  {{\frac{\partial }{\partial\tau}}}W_{ib}^a & = \nabla_i\left(R^a_cJ^c_b - R^c_bJ^a_c\right) - {{\frac{\partial }{\partial\tau}}}\Gamma_{ib}^cJ^a_c + {{\frac{\partial }{\partial\tau}}}\Gamma_{ic}^aJ^c_b\\
  &= J^a_cg^{cd}\left(\nabla_dR_{bi} - \nabla_bR_{di}\right)
     + J^c_bg^{ad}\left(\nabla_c R_{di} - \nabla_dR_{ci}\right)
     - R^a_cW_{ib}^c + R_b^cW_{ic}^a\\
  &= -g^{ac}g^{lm}\nabla_lR_{mibd}J^d_c + g^{ad}g^{lm}\nabla_lR_{midc}J^c_b
    - R^a_cW_{ib}^c + R_b^cW_{ic}^a\\
  &= g^{ad}g^{lm}\left(\nabla_lR_{micb}J^c_d + \nabla_lR_{midc}J^c_b\right)
  - R^a_cW_{ib}^c + R_b^cW_{ic}^a\\
  &= \frac{1}{2}g^{ad}g^{lm}({\operatorname{S}}\circ {\operatorname{F}})_{lmibd}- R^a_cW_{ib}^c + R_b^cW_{ic}^a\\
  &= - R^a_cW_{ib}^c + R_b^cW_{ic}^a
 \end{aligned}$$ so that the components of $W$ in the fiber 