# Almost complex manifold fibered by holomorphic sub-manifolds

Suppose $p:(M,J)\rightarrow (N,I)$ is a submersion between smooth manifolds M and N such that:

1. $(M,J)$ is an almost-complex manifold.

2. $(N,I)$ is a complex manifold where $I$ is the integrable almost-complex structure.

3. $p$ is almost-holomorphic, that is $dp\circ J=I\circ dp.$
4. The fibers of $p,$ which are $J$-holomorphic by the previous condition, are holomorphic in the usual sense. That is, the almost complex structure $J$ restricted to any fiber $p^{-1}(x)$ for $x\in M$ is integrable.

Is $J$ integrable?

I believe that the answer to this question is likely to be no, though I am having trouble cooking up a counter-example. The only immediate consequence I see is that the Nijenhuis tensor on $M$ must take values in vertical vector fields. I am primarily interested in the case where $M$ and $N$ are not compact so standard techniques of deformation theory of compact complex manifold don't easily apply, though I don't know the answer in the case when $p$ is proper either. My google searches haven't turned up anything useful yet, so I'd be very interested to hear about any additional conditions one could add to this list, namely additional structure on $M,$ which guarantees that $J$ is integrable if, as I suspect, the answer to my original question is negative. Thanks in advance for any comments.

• Marko's example is good, but I'll just point out that a very common situation that meets all of your criteria is when $N$ is a complex manifold, $M$ is a complex vector bundle over $N$, and $\nabla$ is a compatible connection on $M$ whose $(0,2)$-curvature is nonzero. Then $\nabla$ induces an almost complex structure on $M$ that is not integrable but that meets all $4$ of your criteria. Of course, if the curvature of $\nabla$ has vanishing $(0,2)$-piece, then $M$ does inherit a natural (integrable) holomorphic structure. Maybe this is the kind of extra assumption you seek. – Robert Bryant Sep 17 '15 at 8:51

If I am not reading something wrong, or did some silly calculation mistake, than the answer to your question is no. Let us take $M=\mathbb R^4$ and $N=\mathbb R^2=\mathbb C$ and the map $p$ just the projection $(x_1,y_1,x_2,y_2)\mapsto (x_1,y_1)$. We can choose $J=\left[\begin{matrix} I & A\\0 & I \end{matrix}\right]$, where $I$ is the standard complex structure on $\mathbb R^2$, just as long as $A$ is of the form $\left[\begin{matrix} a & b\\b & -a \end{matrix}\right]$. Almost any nonconstant $A$ should do the trick. For example, for $A=\left[\begin{matrix} 0 & x_1\\x_1 & 0 \end{matrix}\right]$, you can check that $N_J(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2})=-\frac{\partial}{\partial x_1}$.