A question on the twistor space of a manifold

Question

Let $M$ be either (a) self-dual conformal 4-manifold, or (b) hypercomplex $4n$-manifold. In either case one can construct the twistor space $Z$ (in the case (b) $Z=\mathbb{C}\mathbb{P}^1\times M$ as a smooth manifold) which admits a natural structure of a complex analytic manifold. Let $D(Z)$ denote the Douady space of $Z$, though we will be interested only in the space of rational curves in $Z$. $D(Z)$ is a complex analytic space.

Let $p\colon Z\to M$ be the natural smooth map (in case (b) $p$ is the obvious projection). The fibers of $p$ are complex curves isomorphic to $\mathbb{C}\mathbb{P}^1$. Consider the map $q\colon M\to D(Z)$ defined by $q(x)=p^{-1}(x)$. It is well known in the literature (and uses a Kodaira theorem) that the image $q(M)$ is contained in the smooth part $U$ of $D(Z)$. I need a reference to the following fact which seems to be well known to experts:

The map $q\colon M\to U$ is an infinitely differentiable map of smooth manifolds.

The earliest mentioning of this fact in literature I was able to find is in the paper

Atiyah, M. F.; Hitchin, N. J.; Singer, I. M. Self-duality in four-dimensional Riemannian geometry. Proc. Roy. Soc. London Ser. A 362 (1978).

This paper treats only the case (a) (while I need (b)) and states the result without proof in a somewhat different language (see p. 438).

Answer 1

I think that this follows from basic properties of the moduli space $D(Z)$, since one knows that, for each $x\in M$, the normal bundle $\nu_x$ in $Z$ of each fiber $q(x) = p^{-1}(x)\simeq \mathbb{CP}^1$ is isomorphic to $\mathcal{O}(1)^{2n}$.

Specifically, since one then has $H^1\bigl(q(x),\mathcal{O}(1)^{2n}\bigr) = (0)$ for each $x\in M$, the deformation space is unobstructed (by Kuranishi) so this gives that $$ T_{q(x)}D(Z) = \Gamma\bigl(q(x),\nu_x\bigr) \simeq H^0\bigl(q(x),\mathcal{O}(1)^{2n}\bigr) \simeq \mathbb{C}^{4n}. $$

Now consider the map $Q:Z\to D(Z)$ defined by $Q(z) = q(p(z))$. I think you'll agree that $Q$ is smooth, as this holds for any complex manifold $Z$ that has a locally trivial fibration by compact complex submanifolds each of which is unobstructed. (Probably you can find a proof of this general fact in Morrow and Kodaira's Complex Manifolds, which I don't have with me.)

To prove that $q$ (which is injective) is smooth, it will be enough to show that $Q$ is a submersion onto its image in $D(Z)$. For this, you need to compute the differential $Q':TZ\to TD(Z)$ and show that $Q'(z):T_zZ\to T_{Q(z)}D(Z)$ has (real) rank $4n$ everywhere. However, this differential is easy to compute: Choose a (smooth, not holomorphic) splitting $TZ = K \oplus P$, where $K\subset TZ$ is complex line bundle that is the kernel of $p'$, i.e., the tangent vectors to the fibers of $p$. Now, for each $v\in T_zZ$, let $X_v$ be the vector field along the fiber $Q(z) = q(p(z))$ such that $p'(X_v) = p'(z)(v)$ and $X_v(y)$ lies in $P_y$ for all $y\in Q(z)$. Then $X_v$, when regarded as a section of $\nu_{p(z)}$ in the natural way, is holomorphic. Thus, one has the formula $$ Q'(z)(v) = [X_v]\in \Gamma\bigl(q(x),\nu_x\bigr)= T_{Q(z)}D(Z). $$ Moreover, it is clear that $Q'(z)(v) = 0$ if and only if $v$ lies in $K_z$.

Thus, $Q$ is a submersion onto its image and the rest follows by elementary differential topology, since $p:Z\to M$ is also a submersion with the same fibers as $Q$.