# Why does a deformation modify the complex structure *holomorphically*?

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.

• 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). Jul 25, 2021 at 6:30
• @VeryConfused Is that not just the map $\phi$ being holomorphic? Jul 25, 2021 at 8:54