# Almost complex structure corresponds to unique complex structure up to biholomorphism [on hold]

Assume I have a smooth manifold which has two different holomorphic atlases that induce the same almost complex structure on it. How to show that two complex manifolds (corresponding to the two atlases) are biholomorphic?

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• What do you mean by "the same almost complex structure"? A holomorphic atlas on a smooth manifold $M$ induces a field of endomorphisms $J$ of $T_M$ which is an integrable almost complex structure by definition. Now, if two atlases induce literally the same $J$ than the identity of $M$ is the required biholomorphism. If you mean the two almost complex structures $J_1,J_2$ are isomorphic, then there is $\varphi:M\to M$ such that $d\varphi\circ J_1=J_2\circ d\varphi$, and then again $\varphi$ is the required biholomorphism. – Qfwfq Dec 7 at 18:15
• (to put it succintly, the category of holomorphic manifolds is a full subcategory of that of almost complex manifolds) – Qfwfq Dec 7 at 18:20

Okay, so the focus was on "how to show" rather than whether it's true.

You have a diffeomorphism $$\varphi:M\to M$$ which is an isomorphism of the almost complex manifolds $$(M,J_1)$$ and $$(M,J_2)$$ induced respectively by holomorphic atlases $$\mathcal{A}_1$$ and $$\mathcal{A}_2$$. This means, by definition, that the differential obeys $$d\varphi\circ J_1=J_2\circ d\varphi$$. You want to show that $$\varphi$$ is a biholomorphism of the holomorphic structures on $$M$$ given respectively by $$\mathcal{A}_1$$ and $$\mathcal{A}_2$$.

This follows from the fact that a smooth map $$\phi: U\to V$$, where $$U\subseteq \mathbb{C}^n$$ and $$V\subseteq \mathbb{C}^m$$ are open, is holomorphic if and only if its differential is $$\mathbb{C}$$-linear. This is explained, if I remember well, in the first chapter of Fritzsche--Grauert, From Holomorphic Functions to Complex Manifolds. You'll also use the smooth inverse function theorem, and the fact that a linear map of (real) vector spaces is $$\mathbb{C}$$-linear iff it can be put in the block form

$$\left(\begin{array}{cc} A & -B\\ B & A \end{array}\right)$$

and if it's invertible, also its inverse has the above form.

Lemma: Let $$F \colon M \to N$$ be a smooth map between complex manifold, and let $$I$$ and $$J$$ stand for the almost complex structures on $$M$$ and $$N$$, respectively (considered as smooth manifolds). Then $$F$$ is holomorphic (in the usual sense, i.e. it looks as a holomorphic map in local holomorphic coordinates on $$M$$ and $$N$$) if and only if it respects the complex structures, that is, if $$\forall \; p \in M \;\;\, dF_p \colon T_p M \to T_{F(p)} N$$ is $$\mathbb{C}$$-linear with respect to $$I_p$$ and $$J_{F(p)}$$.

Proof: Observe that if you have a complex $$n$$-manifold, then in local holomorphic coordinates its almost complex structure is just the multiplication by $$i$$ in (the tangent spaces to) $$\mathbb{C}^n$$ (essentially by definition). Therefore, if you pass to local holomorphic coordinates, the lemma will reduce to the following: a smooth complex-valued function $$f$$ on an open subset $$U \subseteq \mathbb{C}^n$$ is holomorphic if and only if $$\forall \, p \in U \;\;\, df_p \colon T_p U \cong \mathbb{C}^n \to T_{f(p)} \mathbb{C} \cong \mathbb{C}$$ is $$\mathbb{C}$$-linear. But then we are done, because this is one of the definitions of holomorphicity, isn't it? (One can easily check that this is equivalent to Cauchy-Riemann) $$\quad \square$$

Now, for your question, just take $$M$$ to be your manifold with the complex structure given by the first atlas, $$N$$ by the second, and $$F$$ the identity map.

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Ivan Solonenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

I think that this follows from the existence of local holomorphic coordinates. Since the induced a.c.s's are the same, we get that in a sufficiently small neighbourhood of any point local holomorphic coordinates can be taken to be the same (it should be obvious how to define the necessary biholomorphism from here).

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• The biholomorphism is the identity map. The two atlases are compatible. – Mike Miller Dec 7 at 18:07
• @MikeMiller I think the real question of OP is whether two definitions of biholomorphism (one similar to the definition of smooth map between smooth manifolds, another is via commutation of differential with almost complex structures) are equivalent. My post only explains why the identity (which obviously is a biholomorphism according to the second definition) is also a biholomorphism according to the first---in small neighbourhoods, it's a composition of biholomorphism to an open ball and of its inverse. If you think that can be shown in a simpler way, I would gladly look at your proof. – rori Dec 7 at 18:35