# Pushing forward a complex structure by submersion

I have a surjective smooth map with surjective differential between two balls $$\phi:B^{2n}\rightarrow B^{2k}$$. Fix an integrable almost complex structure $$J$$ on $$B^{2n}$$. Assume that $$\mathrm{Ker}\:d\phi$$ is preserved by the action of $$J$$.

For any point $$q \in B^{2k}$$, I can find a point $$p\in B^{2n}$$ satisfying $$\phi(p)=q$$ and I can pushforward the complex structure from $$T_{B^{2n}, p}$$ to $$T_{B^{2k}, q}$$. Is it true that the resulting complex structure on $$T_{B^{2k}, q}$$ does not depend on the lift? If so, do I get an integrable almost complex structure on $$B^{2k}$$?

I think that the answer to the first question is positive if $$\phi^{-1}(q)$$ is connected as for sufficiently small open sets in the fiber we have local normal form (and independence from the lift becomes self-evident). However, as Mike Miller mentions, the fibers don't have to be connected.

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• No. You will need some properness condition to avoid many counterexamples. Imagine a small ball snaking around the unit ball until it has covered everything, and consider this as a map from $(-t,t) \times D^k \cong D^{k+1}$. It is similarly not hard to come up with a smooth local diffeomorphism $D^2 \to D^2$ so that some fibers are points and some are two using essentially the same idea. – Mike Miller Dec 7 at 8:22

In general, even if the fibers of $$\phi$$ are connected and the map is proper, there need not be an almost complex structure on $$B^{2k}$$ such that the differential of $$\phi$$ is complex linear. (I assume that you meant to assume that the kernel of the differential of $$\phi$$ is preserved by $$J$$, not that $$J$$ acts trivially on it, which doesn't make sense unless the kernel is $$0$$.)

Perhaps the most famous example of this phenomenon is the so-called 'twistor fibration': $$\pi:\mathbb{CP}^3\to \mathbb{HP}^1=S^4$$. The idea is that we regard $$\mathbb{C}^4$$ as $$\mathbb{H}^2$$ (where $$\mathbb{H}$$ is the ring of quaternions). Then $$\mathbb{CP}^3$$ is the space of $$1$$-dimensional subspaces of $$\mathbb{C}^4$$ and the map $$\pi$$ is defined by letting $$\pi(\mathbb{C}{\cdot}v) = \mathbb{H}{\cdot}v$$ for every nonzero $$v\in\mathbb{C}^4$$. The fibers of $$\pi$$ are complex lines in $$\mathbb{CP}^3$$, so they are holomorphic submanifolds of $$\mathbb{CP}^3$$ that are connected and compact, and $$\pi$$ is a submersion. However, $$S^4$$ does not admit any almost-complex structure at all, let alone one for which the differential of $$\pi$$ is $$\mathbb{C}$$-linear. In fact, it's not hard to see that this is a local failure, in the sense that, even after restricting $$\pi$$ to some (non-empty) open subset $$U\subset\mathbb{CP}^3$$, there still is no almost-complex structure on the image $$\pi(U)$$ for which the differential of $$\pi$$ is $$\mathbb{C}$$-linear.

There are many other examples, of course, even an example of a submersion $$\phi:B^4\to B^2$$, where $$B^{2k}\subset\mathbb{C}^k$$ is the usual open ball, such that the fibers of $$\phi$$ are connected complex curves in $$B^4$$ but there is no almost-complex structure on $$B^2$$ for which the differential of $$\phi$$ is $$\mathbb{C}$$-linear. I can supply this example, too, if the OP is interested.

Meanwhile, if there is an almost complex structure on $$B^{2k}$$ for which the differential of $$\phi$$ is complex linear, then, yes, that almost complex structure has to be integrable.

The proof of this is straightforward: Suppose that there is an almost-complex structure on $$B^{2k}$$ for which the differential of $$\phi$$ is $$\mathbb{C}$$-linear. Let $$\omega^1,\ldots,\omega^k$$ be a $$C^\infty(B^{2k})$$ basis for the $$(1,0)$$-forms on $$B^{2k}$$ with respect to this almost-complex structure. Then there will be functions $$N^i_{\overline{p}\overline{q}}=-N^i_{\overline{p}\overline{q}}$$ on $$B^{2k}$$ such that the equations $$\mathrm{d}\omega^i \equiv \tfrac12\,N^i_{\overline{p}\overline{q}}\, \overline{\omega^p}\wedge\overline{\omega^q} \quad\mathrm{mod}\quad \omega^1,\ldots,\omega^k$$ for $$1\le i\le k$$ hold. These functions vanish identically if and only if the almost-complex structure is integrable. Now, let $$^*\!\omega^i=\phi^*(\omega^i)$$. Because the differential of $$\phi$$ is $$\mathbb{C}$$-linear and $$\phi$$ is a submersion, these complex-valued $$1$$-forms are linearly independent and of type $$(1,0)$$ with respect to the $$J$$ on $$B^{2n}$$. Thus, they can be completed to a $$C^\infty(B^{2n})$$-basis of $$(1,0)$$-forms for $$J$$ by choosing some $$(1,0)$$-forms $$^*\!\omega^{k+1},\ldots,^*\!\omega^n$$ on $$B^{2n}$$ such that $$^*\!\omega^1,\ldots,^*\!\omega^n$$ are linearly independent. However, because $$J$$ is integrable, we must have $$\mathrm{d}(^*\!\omega^j)\equiv 0\quad\mathrm{mod}\quad ^*\!\omega^1,\ldots,^*\!\omega^n$$ for $$1\le j\le n$$. This immediately implies that $$\phi^*(N^i_{\overline{p}\overline{q}})=0$$, and, since $$\phi$$ is a surjective submersion, this implies that $$N^i_{\overline{p}\overline{q}} =0$$, i.e., that the 'push-forward' almost-complex structure is integrable.

• Do you have a reference or an explicit example at hand? I am asking because theorem 2.26 in volume 1 of Voisin's text says "Let X be a complex manifold of dimension n, and let E be a holomorphic distribution of rank k over X, i.e. a holomorphic vector subbundle of rank k of the holomorphic tangent bundle TX . Then E is integrable in the holomorphic sense if and only if we have the integrability condition $[E, E]=E$." To prove it, she considers the real part of $E$, applies Frobenius theorem to it so she gets a submersion $U\rightarrow V$ as in the question... – lolo Dec 7 at 10:04
• Then she claims that there exists an almost complex structure on $V$ in which the differential becomes $\mathbb{C}$-linear. – lolo Dec 7 at 10:04
• and yes I made a stupid mistake in the question, of course $J$ can not fix anything but zero vector – lolo Dec 7 at 10:04
• @lolo: Yes, I'll edit my answer to include the example and the argument for the second question. I'm sorry that I didn't do it first, but I had to stop after I put in my answer and tend to some other business. – Robert Bryant Dec 7 at 12:40
• @lolo: By the way, I forgot to respond to your worry about the apparent conflict of such examples with Voisin's text. The reason that there isn't a contradiction is that she is assuming that $E$ is a holomorphic subbundle of $TX$, whereas you have only assumed that $\mathrm{ker}\,\mathrm{d}\phi$ is a complex subbundle of $TX$ (one that is also, of course, Frobenius as a (real) subbundle of $TX$, but that doesn't make it holomorphic). – Robert Bryant Dec 7 at 17:21