# Almost complex manifold of dimension 2… locally isomorphic to ℂ?

I know that this is supposed to be standard, but I don't know how to search for it... hence the question:

Let $$J$$ be an almost complex structure on $$M:=\mathbb R^2$$, i.e., a $$C^\infty$$ section of $$\mathrm{End}(TM)$$ which squares to $$-\mathrm{id}_{TM}$$.

How does one prove that $$(M,J)$$ is integrable, i.e. how does one prove that it is locally isomorphic to $$\mathbb C$$?

• Hi this absolutely isn’t my field but... I think one can show the vanishing of the Nijenhuis tensor easily: N_J(X, X) = 0 and N_J(X, JX) = 0 by inspection. Let me know if I’ve said something dumb (true with high probability)! – alpoge May 18 '19 at 1:05
• If you can find a vector field $X$ on a neighborhood U such that $[X,JX]=0$ on $U$, then that neighborhood is isomorphic to $\mathbb{C}$. This is because the Lie bracket is the obstruction to $X,JX$ "integrating" to a coordinate grid – Daniel Barter May 18 '19 at 1:28
• – Raju May 18 '19 at 8:02

Eckmann-Frölicher say so right after they introduce the (“Nijenhuis”) torsion tensor in (1951, p. 2284), bottom: “dans $$\mathrm E_2$$, $$t^j_{kl}$$ est toujours $$=0$$.”
For $$n=1$$ the compatibility conditions are vacuous and the problem becomes that of introducing isothermal coordinates with respect to the Riemannian metric $$ds^2=|dz + ad\bar z|^2$$