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
up vote 2 down vote accepted

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

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