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Let's say we have an almost complex manifold $(M, J)$. Consider the complex vector bundle $V\rightarrow M$ whose fiber over $x$ is the space of almost complex structures on $T_x M$.

Is there any logical connection between the following two conditions:

  • $J$ is integrable;
  • the map $M\rightarrow V$ associated to $J$ is holomorphic?
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    $\begingroup$ "Consider the complex vector bundle $V\to M$ ..." - In which way do you see $V\to M$ as a vector bundle? The condition $J^2=-1$ is not linear in $J$. $\endgroup$
    – Qfwfq
    Commented Jan 6, 2019 at 22:31

2 Answers 2

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First of all, there is no reasonable complex structure on the total space of this bundle — surely, it should be the usual complex structure on the space $\mathrm{SO}(2n)/\mathrm{U}(n)$ along the fibres; and it would be natural to extend it tautologically in the horizontal direction, i. e. for $(\iota,x)\in\mathcal{J}(M)$ a complex structure operator on the tangent space $T_xM$, one would have $I(v) = \widetilde{\iota(d\pi(v))}$, where $\pi \colon \mathcal{J}(M) \to M$ is the projection and $v \in T_{(\iota, x)}\mathcal{J}(M)$ is a horizontal vector. The issue is that there is no natural notion of a “horizontal vector” unless you pick up a connection in the bundle $\mathrm{End}(TM)$. However, if your section is horizontal w. r. t. a connection in the bundle of endomorphisms which comes from a torsion free connection in the tangent bundle $TM$, then your almost complex structure would be integrable according to a well-known theorem in differential geometry, which is discussed e. g. in the great MO post on the geometric meaning of torsion.

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    $\begingroup$ (A minor remark: the bundle with fiber $\mathrm{SO}(2n)/\mathrm{U}(n)$ parametrizes almost complex structures which are orthogonal with respect to a given Riemanniann metric and preserve a given orientation on $M$, rather then all almost complex structures) $\endgroup$
    – Qfwfq
    Commented Dec 8, 2018 at 1:50
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    $\begingroup$ but if $M$ is an open ball with a fixed embedding to $\mathbb{R}^{2n}$, then the restriction of Levi--Civita connection from $\mathbb{R}^{2n}$ provides us with a flat connection on $TM$. Do we get a complex structure on the total space of the bundle? $\endgroup$
    – rori
    Commented Dec 8, 2018 at 5:26
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This is an answer to what was perhaps the intended question (taking into account the comment to the answer by @gulag57). Let $U$ be an open set in $\mathbb{C}^n$. An almost complex structure on $U$ can be regarded as a map $J \colon U \to \text{GL}(2n;\mathbb R)/\text{GL}(n;\mathbb C)$. The target space has a canonical structure of complex manifold, as is elegantly explained in section 1 of this Bourbaki seminar by Douady. One may then ask whether complex structures on $U$ correspond to maps $J$ which are holomorphic. That is not the case.

Almost complex structures which are sufficiently close to the standard complex structure on $U$ are parametrized by maps $\theta \colon U \to \text{Hom}( \overline{ \mathbb C^n}, \mathbb C^n)$ (there are canonical complex charts around each point in $\text{GL}(2n;\mathbb R)/\text{GL}(n;\mathbb C)$ taking values in $\text{Hom}( \overline{ \mathbb C^n}, \mathbb C^n)$ as is also explained in Douady's paper). The map $\theta$ will correspond to an integrable structure if and only if it satisfies the Maurer-Cartan equation $\overline{\partial} \theta + [\theta,\theta]=0$ (see for instance "Complex Geometry" by Daniel Huybrechts, Lemma 6.1.2 p. 258). This is very different from the map $\theta$ being holomorphic.

For instance, the anti-holomorphic map defined by $$\theta(z_1,z_2) = \overline{z_2} d\overline{z_1} \otimes \frac{\partial}{\partial z_1} + \overline{z_1} d\overline{z_2} \otimes \frac{\partial}{\partial z_1}$$ or, in matrix notation, $$ \theta(z_1,z_2) = \left[ \begin{array}{cc} \overline{z_2} & \overline{z_1} \\ 0 & 0 \end{array} \right] $$ satisfies the Maurer-Cartan equation and therefore corresponds to a complex structure on $\mathbb C^2$.

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