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Given a Kähler manifold $\mathcal{N}$ and an arbitrary (i.e. non-Kähler) differentiable sub manifold $\mathcal{M}\subset\mathcal{N}$, I would like to foliate $\mathcal{M}$ with maximal Kähler sub manifolds.

Intuitively, I can restrict the complex structure field $J$ from the Kähler manifold $\mathcal{N}$ using the orthogonal projection (using the Riemannian metric on $\mathcal{M}$). At every point $p$, there will be a maximal subspace $\overline{T_{p}\mathcal{M}}\subset T_{p}\mathcal{M}$ on which $J_{\mathrm{res}}^2=-1$. The choices of $\overline{T_{p}\mathcal{M}}$ forms a geometric distribution on $T\mathcal{M}$.

I would expect that I can typically (or always?) integrate the geometric distribution to find a collection of sub manifolds $\overline{\mathcal{M}}\subset\mathcal{M}$ that are all Kähler manifolds. I expect that this is standard and well-understood, but I did not find a discussion in the standard literature (which I'm not super familiar with).

It would be wonderful if somebody more familiar with the topic, could confirm (or correct) my suspicion and possibly refer me to a standard discussion.

LFH
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    There are complex tori of dimension greater than one whose only submanifolds are points. Let $N$ be such a torus and take $M = N$, then, if I have understood you correctly, you do not obtain what you hoped for. – Michael Albanese Oct 23 '19 at 12:40
  • @MichaelAlbanese: Thank you so much for your reply. Are such tori Kähler manifolds? It's important in my assumption that $\mathcal{N}$ is Kähler. Clearly, for $\mathcal{M}=\mathcal{N}$ all tangent spaces of $T_p\mathcal{M}$ are Kähler spaces, so integrating the respective geometric distribution will just give back $\mathcal{N}=\mathcal{M}$ or did I miss something? – LFH Oct 23 '19 at 12:50
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    Yes they are. All tori admit Kähler metrics. Well, any manifold can be foliated by itself, or by points. I was assuming you were looking for a non-trivial foliation. – Michael Albanese Oct 23 '19 at 12:57
  • Ah, I see. Well, I just want to foliate by Kähler manifold. If the submanifold $\mathcal{M}$ is already Kähler, the situation is trivial, because I can foliate it by itself. However, I'm interested in scenarios where $\mathcal{M}$ is NOT Kähler (only $\mathcal{N}$). In this case, my construction will not give rise to a self-foliation. – LFH Oct 23 '19 at 13:04
  • Do you want $N$ to be arbitrary or do you require it to be non-Kähler? Also, is $N$ a complex submanifold? – Michael Albanese Oct 23 '19 at 13:15
  • Thanks for your question. $\mathcal{N}$ is the large manifold and always Kähler. The sub manifold $\mathcal{M}$ should be arbitrary, but the setup is trivial if $\mathcal{M}$ is Kähler. Consequently, we should take $\mathcal{M}$ to be non-Kähler and also not a complex sub manifold, because any complex sub manifold of a Kähler manifold is automatically a Kähler manifold itself (which would go back to the trivial case). – LFH Oct 23 '19 at 13:44
  • So what about the case when $N$ is a complex torus of dimension 2 of the type I mentioned and $M$ is a real three-torus? – Michael Albanese Oct 23 '19 at 16:20
  • What happens in the scenario that you are describing. How large are the smallest complex subspaces of the tangent spaces of the 3-torus? – LFH Oct 23 '19 at 20:37

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I found the following reference: https://www.researchgate.net/publication/227124990_Differential_geometry_of_real_submanifolds_in_a_Kaehler_manifold.

LFH
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