This is a question regarding Chapter 9.1 of Claire Voisin's book [1]

Let $\phi: \mathcal X \to B$ be a family of compact complex manifolds, that is a proper holomorphic submersion, with central fiber $X = X_0 = \phi^{-1}(0)$. Then by Ehresmann's theorem we can find a $\mathcal C^\infty$-trivialization $$ T=(\pi, \phi): \mathcal X \xrightarrow{\cong} X \times B. $$ According to Proposition 9.5 in [1], we can even choose $T$ in such a way, that the fibers of the compositie map $$ \pi: \mathcal X \to X \times B \xrightarrow{\operatorname{pr}_X} X$$ are complex submanifolds of $\mathcal X$. Voisin then writes

These fibres are submanifolds of $\mathcal X$ which are diffeomorphic to $B$. The fact that they are comlex does not, of course, imply that $\pi$ is holomorphic, since these submanifolds do not vary holomorphically with the point $x \in X_0$, but it does say that the family of complex structures on $X_0$ parametrised by $B$ [...] varies holomorphically with $t \in B$.

Q: What exactly does it mean that the family of complex structures "varies holomorphically"?

A bit further down, Voisin seems to specify this notion, but she doesn't really give an argument and I'm not really convinced what she writes is true.

We can identify deformations $X_t$ with $X$ via $T$, so we get for each $t$ a decomposition of the complexified tangent bundle $$T_{X, \mathbb R} \otimes_{\mathbb R} \mathbb C = T_{X, \mathbb C} = T_t^{0,1} \oplus T_t^{1,0}.$$ If $t$ is small, the projection $T_{X, \mathbb C} \to T^{0,1}$ induces isomorphisms $f: T_t^{0,1} \cong T^{0,1}$, and using this we can define a map $$\alpha_t = -(\operatorname{pr}_{T^{1,0}} \circ f^{-1}): T^{0,1} \to T^{1,0},$$ which has the property that $$T^{0,1}_t = \{u - \alpha_t(u) \,|\, u \in T^{0,1}\}.$$ Again, Voisin claims that $\alpha_t$ varies holomorphically with $t \in B$. The proof of the following Proposition 9.7 hints at how to compute (at least locally) the derivatives $\frac{\partial \alpha_t}{\partial \overline{t_k}}$, so let's see what happens there and if we can verify the holomorphicity condition $$\frac{\partial \alpha_t}{\partial \overline{t_k}} = 0. $$

Take a point $p \in X_0 \subset \mathcal X$. Since $\phi$ is a holomorphic submersion, we can choose local coordinates on $z_1, \dotsc, z_n, t_1, \dotsc, t_r$ on $\mathcal X$, such that $\phi$ is given by $$(z_1, \dotsc, z_n, t_1, \dotsc, t_r) \mapsto (t_1, \dotsc, t_r).$$ For local coordinates $w_1, \dotsc, w_n$ on $X$, the map $\pi$ is given by the components $$\pi(z,t) = (\pi_1(z,t), \dotsc, \pi_n(z,t)).$$ From $\pi_*\left(\frac{\partial}{\partial \overline{z_i}}\right) = \sum_j \frac{\partial \pi_j}{\partial \overline{z_i}}\frac{\partial}{\partial z_j} + \sum_j \frac{\partial \overline{\pi_j}}{\partial \overline{z_i}}\frac{\partial}{\partial \overline{z_j}}$ it follows by definition of $\alpha_t$, that $$\alpha_t\left(\sum_j \frac{\partial \overline{\pi_j}}{\partial \overline{z_i}}\frac{\partial}{\partial \overline{z_j}} \right) = - \sum_j \frac{\partial \pi_j}{\partial \overline{z_i}}\frac{\partial}{\partial z_j}.$$ Applying $\frac{\partial}{\partial \overline{t_k}}$ we get $$\sum_j \frac{\partial^2 \overline{\pi_j}}{\partial \overline{t_k} \partial \overline{z_i}} \cdot \alpha_t\left(\frac{\partial}{\partial z_j}\right) + \sum_j \frac{\partial\overline{\pi_i}}{\partial \overline{z_j}} \cdot \frac{\partial \alpha_t}{\partial \overline{t_k}}\left(\frac{\partial}{\partial \overline{z_j}}\right) = - \sum_j \frac{\partial^2 \pi_j}{\partial \overline{t_k} \partial z_j} \cdot \frac{\partial}{\partial z_j}$$ Let's now restrict to $t = 0$. Then $\alpha_0 = 0$, and $\pi = \operatorname{id}_X$. Hence the first sum disappears completely, and from the second sum we only get $$\frac{\partial \alpha_t}{\partial \overline{t_k}}|_{t=0}\left(\frac{\partial}{\partial z_i} \right) = - \sum_j \frac{\partial^2 \pi_j}{\partial \overline{t_k}\partial \overline{z_i}} \cdot \frac{\partial}{\partial z_j}$$

At this point I don't know why the right hand side should vanish. Voisin claims, that the $\pi_j$ are holomorphic in the $t_k$, which would be sufficient. However, I think this need not be true. For example we could have $n = r = 1$ and $\pi(z,t) = \overline{t} - \overline{z}$. Then $\pi$ is not holomorphic in $t$, but its fibers $$\pi^{-1}(c) = \{(z,t) \,|\, t - z = \overline{c} \}$$ are still complex submanifolds of $\mathbb C^2$.

[1] Claire Voisin, Hodge Theory and Complex Algebraic Geometry, I.

  • $\begingroup$ Regarding your concern about "varying homomorphically", consider the following example: Let $f : X \to \mathbb{P}^1$ be a family of elliptic curves (i.e., complex tori $\mathbb{C} / \Lambda$). The complex structure of these elliptic curves is parametrised by the $j$--invariant. Hence, the complex structure on the fibres varies homomorphically if the $j$--invariant is holomorphic (which is the case). $\endgroup$
    – AmorFati
    Jul 25, 2021 at 6:30
  • $\begingroup$ @VeryConfused Is that not just the map $\phi$ being holomorphic? $\endgroup$ Jul 25, 2021 at 8:54


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