2
$\begingroup$

Show that any almost complex structure is induced by at most one complex structure.

Suppose $X$ is a complex manifold of dimension $n$ and $M$ is the underlying $2n$ dimensional real manifold.

If $J_1$ and $J_2$ are two almost complex structures induced by the same complex structure on $X$ I would want to show that $J_1 = J_2$.

I know that any almost complex structure induced by an complex structure is integrable so for both $J_1$ and $J_2$ we have that the Nijenhuis tensor $$N(X,Y)=[X,Y]+J([JX,Y] + [X,JY]) - [JX,JY]$$ vanishes. I'm wondering on whether this is enough to conclude that $J_1=J_2$?

$\endgroup$
6
  • $\begingroup$ But that is not what you want to prove. That would prove that every complex structure induces a unique almost complex structure (do you think that is true?). But you want to show that if two complex manifolds induce the same almost complex structure, then the complex manifolds are isomorphic. This is a different statement. $\endgroup$ Commented Aug 17, 2023 at 11:49
  • $\begingroup$ Ah I had some misunderstanding here. So is the correct statement to consider two complex manifolds $X_1$ and $X_2$ with the same underlying real manifold $M$ such that $X_1$ and $X_2$ induce almost complex structures $J_1$, $J_2$ with $J_1 = J_2$ and try to show that $X_1 \cong X_2$? @QuaereVerum $\endgroup$
    – Danlo
    Commented Aug 17, 2023 at 11:58
  • $\begingroup$ Yes, that's what would prove the statement in the title of this post. $\endgroup$ Commented Aug 17, 2023 at 12:01
  • $\begingroup$ Thanks for clarifying this. I couldn't find a definition for a complex structure, but I assume it is just an holomorphic atlas $\{\varphi_\alpha : U_\alpha \to \mathbb{C}^n \}$ so in light of the problem I could consider the two complex atlases $\{\psi_\beta : V_\beta \to \mathbb{C}^n \}$, $\{\varphi_\alpha : U_\alpha \to \mathbb{C}^n \}$ and it would be sufficient to show that these two give the same maximal atlas? @QuaereVerum $\endgroup$
    – Danlo
    Commented Aug 17, 2023 at 12:07
  • $\begingroup$ What you want to show is: given two complex manifolds with the same underlying smooth manifold, if the bundle map $J:TM\to TM$ that they induce is the same, then the complex manifolds are isomorphic. This bundle map is defined in local holomorphic coordinates $(z^j=x^j+iy^j)$ by $J(\frac{\partial}{\partial x^j})=\frac{\partial}{\partial y^j}$ and $J(\frac{\partial}{\partial y^j})=-\frac{\partial}{\partial x^j}$. $\endgroup$ Commented Aug 17, 2023 at 12:18

1 Answer 1

4
$\begingroup$

To address the problem in the title, rather than the problem you set up in the subsequent text: let $\mathcal{X}_1$ and $\mathcal{X}_2$ denote two complex structures on an underlying manifold $X$. There are corresponding bundle maps $J_1,J_2:TX\to TX$. Suppose that $J_1=J_2$, or (more generally) that $f^*J_1=J_2$ for some diffeomorphism $f:X\to X$. Then we must show that $\mathcal{X}_1\cong\mathcal{X}_2$, as complex manifolds. This amounts to checking that $f$ (which in the simplest case is $f=\text{id}$) is a biholomorphism. Now prove the following:

Lemma. Let $f:\mathcal{X}\to\mathcal{Y}$ be a smooth map between complex manifolds with induced almost complex structures $J_X$ and $J_Y$. Then $f$ is holomorphic if and only if $df\circ J_X=J_Y\circ df$.

Using the lemma, you can now show that the assumption that $f^*J_1=J_2$ implies that $f$ must be a biholomorphism, and therefore $\mathcal{X}_1\cong\mathcal{X}_2$.

$\endgroup$
2
  • $\begingroup$ Isn't $f:\mathcal{X}\to\mathcal{Y}$ actually the identity here if the manifolds have the same underlying smooth manifold? Or at least we could pick it as a candidate for the biholomorphism. @quaere-verum $\endgroup$
    – Danlo
    Commented Aug 18, 2023 at 9:03
  • $\begingroup$ As noted, in this case (which is the simplest case), $f$ is indeed the identity. More generally, it can be a self-diffeomorphism, and this case is also import to consider if one wants to construct the moduli space of complex structures on $X$. However, it is important to be aware of the fact that the identity map on $X$ is NOT by default a biholomorphism. Whether or not it is, depends on the given complex structures. $\endgroup$ Commented Aug 18, 2023 at 11:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .