# An almost complex structure on real 2-dimensional manifold

Why an almost complex structure on real 2-dimensional manifold is integrable?

• Because the obstruction vanishes for dimension reason. – Moishe Kohan Jun 17 '14 at 14:52

With respect to $J$ we get a splitting of the complexified differential forms; namely,

$$\Omega^k(X)\otimes\mathbb{C} = \bigoplus_{p+q=k}\Omega^{p,q}(X).$$

With respect to this splitting, we have the operator $\overline{\partial} : \Omega^{p,q}(X) \to \Omega^{p,q+1}(X)$ given by $\pi^{p,q+1}\circ d$ where $d$ is the exterior derivative and $\pi^{p,q+1}$ is the projection $\Omega^{k+1}(X)\otimes\mathbb{C} \to \Omega^{p,q+1}(X)$.

Integrability of $J$ is equivalent to the condition $\overline{\partial}^2 = 0$, but this is automatic on a real surface because we have

\begin{align*} \Omega^0(X)\otimes\mathbb{C} &= \Omega^{0,0}(X)\\ \Omega^1(X)\otimes\mathbb{C} &= \Omega^{1,0}(X) \oplus \Omega^{0,1}(X)\\ \Omega^2(X)\otimes\mathbb{C} &= \Omega^{1,1}(X) \end{align*}

and $\Omega^{p,q}(X) = 0$ for all other $p$ and $q$. Now note that

$$\Omega^{0,0}(X) \xrightarrow{\overline{\partial}} \Omega^{0,1}(X) \xrightarrow{\overline{\partial}} \Omega^{0,2}(X) = 0.$$

So $\overline{\partial}^2 = 0$ on $\Omega^{0,0}(X)$. Likewise, $\overline{\partial}^2 = 0$ on $\Omega^{1,0}(X)$, $\Omega^{0,1}(X)$, and $\Omega^{1,1}(X)$.

Added Later: Altenatively, we can show that for any almost complex structure $J$ on a smooth two-dimensional manifold, $N_J = 0$ where $N_J$ is the Nijenhuis tensor field of $J$. By the Newlander-Nirenberg Theorem, the vanishing of $N_J$ is equivalent to integrability.

Let $M$ be a smooth manifold with almost complex structure $J$. Recall that the Nijenhuis tensor field of $J$ is given by

$$N_J(X, Y) = [X, Y] + J[JX, Y] + J[X, JY] - [JX, JY].$$

As $N_J$ is tensorial, if $\{V_1, \dots, V_n\}$ is an ordered basis of local vector fields defined in a neighbourhood of $p$, $N_J$ vanishes at $p$ if and only if $N_J(V_i, V_j) = 0$ for all $1 \leq i, j \leq n$. Furthermore, $N_J$ is skew-symmetric, so it is enough to show that $N_J(V_i, V_j) = 0$ for all $1 \leq i < j \leq n$.

Now consider the case where $M$ is a smooth two-dimensional manifold. Fix $p \in M$ and let $V$ be a nowhere zero local vector field defined in a neighbourhood of $p$, then $\{V, JV\}$ is an ordered basis of local vector fields; to see this, note that if $JV \in \operatorname{span}\{V\}$ then $JV = kV$ so $-V = J^2V = k^2V$ which is impossible. As

\begin{align*} N_J(V, JV) &= [V, JV] + J[JV, JV] + J[V, J^2V] - [JV, J^2V]\\ &= [V, JV] + J[V, -V] - [JV, -V]\\ &= [V, JV] - J[V, V] + [JV, V]\\ &= [V, JV] - [V, JV]\\ &= 0, \end{align*}

we see that $N_J = 0$ at $p$. As $p$ is arbitrary, the Nijenhuis tensor field of $J$ vanishes, so $J$ is integrable.

• Wonderful answer,thanks a lot @Michael Albanese. – Steve Sep 22 at 9:53