# Equivalence of (almost) complex structures

Preamble: An almost complex structure on a manifold $M$ is an endomorphism $J : TM \to TM$ such that $J^2 = -1$. An almost complex structure $J$ is said to be integrable if the Nijenhuis tensor, $N_J$, vanishes. $M$ admits an integrable almost complex structure iff $M$ is a complex manifold, i.e. $M$ has an atlas of charts to the open unit disc in $\mathbb{C}^n$ with holomorphic transition functions.

When are two (almost) complex structures considered equivalent?

For example, if $J$ is an (almost) complex structure, then $-J$ is also an (almost) complex structure. In what sense is this a new (almost) complex structure? Indeed, Michael Albanese's answer on this question shows that there can be quite a few (almost) complex structures on a manifold.

A seemingly natural way to define equivalence for complex structures is bilomorphicity - two complex structures are equivalent if the associated complex manifolds are biholomorphic. This seems like a reasonable thing to do, but doesn't work if you have an almost complex structure which is not integrable (since in that case there isn't an associated complex manifold).

• There's no obstruction to you defining biholomorphic anyway. A ($J$- or pseudo-)holomorphic map between almost complex manifolds is just a map $\varphi: M \to N$ such that $d\varphi(J_Mv) = J_N d\varphi(v)$. You can check that this is the same as the standard notion of holomorphic in the case that $M,N$ are complex. – user98602 Aug 18 '15 at 1:29
• There are various notions depending on what you want to do. One notion that arises in algebraic topology is homotopy-theoretic (and preserves e.g. Chern classes), but with respect to this notion every pair of e.g. Riemannian metrics is equivalent and so it's not very geometrically interesting. To get some interesting geometry you probably want to do what Mike Miller suggests. – Qiaochu Yuan Aug 18 '15 at 8:50