7
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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 $Q$.

$\endgroup$
  • $\begingroup$ For me, to agree that $Q$ is infinitely differentiable is as hard (or as easy) as to agree that $q$ itself is infinitely differentiable. Indeed, if $q$ is infinitely differentiable then $Q=q\circ p$ is infinitely differentiable since $p$ is. Conversely, locally on $M$ the map $p$ admits an infinitely smooth section $s\colon M\to Z$ such that $p\circ s=Id$ (since $p$ is a submersion). Then $q=Q\circ s$ is infinitely differentiable. Thus I will greatly appreciate the precise statement and reference to the general fact you mentioned, I will help a lot. $\endgroup$ – orbits Jan 25 '15 at 9:46
  • $\begingroup$ I agree, except that the argument above proves something more, namely that $q$ is not only smooth but embeds $M$ as a submanifold of $D(Z)$. For the general fact about locally trivial foliations by compact complex submanifolds, I referred you to Morrow and Kodaira, where I believe that this general fact (or something that implies it) is proved as part of the Kuranishi deformation space theorem. I am traveling now and don't have access to the book, but I'll check when I get home and give you a more precise reference. $\endgroup$ – Robert Bryant Jan 25 '15 at 10:24
  • $\begingroup$ I finally got home and looked in Morrow and Kodaira, only to find that the theorem I was expecting to find was not there. I'll look at some other possible references in the next day or so when I have time; I'm sure that I have seen the general result that I quoted somewhere, I just have to find it. $\endgroup$ – Robert Bryant Jan 27 '15 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.