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Let $M$ be a closed real-analytic manifold of dimension $2n$. Is it possible to make sense of the moduli space of real-analytic almost complex structures on $M$ as an algebro-geometric object (probably a very non-Noetherian one)? Can this be used to gain a new perspective on, say, Fredholm-regular almost complex structures? I would not think that this would be particularly useful, but I think if this is possible it is worth doing just for the fun of it.

Maybe one terribly non-canonical way to do this is to construct it a subscheme of $\bigcup_{p\in M}\mathrm{Mat}(2n, 2n)$ (choose a basis for the fiber of tangent bundle at some point, choose a connection to produce bases in points nearby, somehow make sense of the real-analyticity condition, and then take the vanishing scheme of the "polynomial" equation $J^2=-\mathrm{Id}$ and hope that this can be made to work globally). Hopefully, somebody else thought of a better way.

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    $\begingroup$ It is the space of analytic sections of the bundle of almost complex structures on the fibers; I don't see the need for any connection, or the need to work locally. $\endgroup$
    – Ben McKay
    Commented Mar 25, 2019 at 8:52
  • $\begingroup$ @BenMcKay and is there a reference where it is described how to make sense of that as a scheme? $\endgroup$
    – user137400
    Commented Mar 25, 2019 at 9:10
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    $\begingroup$ What do you mean by "an algebro-geometric object"? If everything is real-analytic, would you expect the resulting object to be "algebraic"? $\endgroup$
    – Qfwfq
    Commented Apr 28, 2019 at 21:22
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    $\begingroup$ I think the key word could be: twistor space. You can make sense of the bundle of almost-complex structures on the fibers of the tangent bundle of $M$ (called the twistor space of $M$); if $M$ is real-analytic it will likely be a real-analytic bundle with fiber a (real algebraic) homogeneous space. Then the set of (real analytic) almost complex structures on $M$ will be sections of this bundle. Is it a real-analytic object of some sort? I don't know, but sometimes the set of scheme morphisms $X\to Y$ between two schemes has a natural structure of scheme, so something similar could be true. $\endgroup$
    – Qfwfq
    Commented Apr 28, 2019 at 21:28

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Let $M$ be a closed $2n$-dimensional smooth orientable manifold, and let $TM$ be its tangent bundle.

Let $\mathcal{A}(M):=\{J\in C^\infty(M,\mathrm{End}(TM))\ |\ J_x^2=-1,\ \forall x\in M\}$. This is an infinite dimensional smooth manifold in general.

We will say $J,J'\in \mathcal{A}(M)$ are equivalent if there exists a diffeomorphism $f\in \mathrm{Diff}(M)$ such that:

  1. $f_*(J)=J'$, and
  2. there is a continuous map $p:[0,1]\to \mathrm {Diff} (M)$ such that $p(0)=f$ and $p(1)=\mathrm {Id}_M$.

I think this is essentially the conjugation action of $\mathrm{Aut}(TM)$ described here.

Let $\mathcal{J}(M):=\mathcal{A}(M)/\sim$, with respect to the equivalence relation defined above.

Perhaps this is a candidate for a moduli space of almost complex structures on $M$.

When $M=S$ is a surface ($n=1$), all almost complex structures are complex. In that case, the equivalence relation seems to me to be the usual one for Teichmüller space and so $\mathcal{J}(S)=\mathcal{T}(S);$ see here.

As $\mathcal{T}(S)$ covers the moduli space of algebraic curves (of type $S$) via the action of the mapping class group, one could hope that a similar quotient by the general mapping class group $\mathrm{Diff}(M)/\mathrm{Diff}_0(M)$ on $\mathcal{J}(M)$ results in something algebraic.

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