Conformally immersed Riemann surfaces and foliations I want to show that conformally immersed Riemann surfaces in $\mathbb{R}^4$ are leaves of a 2-foliation $\mathcal{F}$. I start with the generalized Weierstrass representation of the surfaces: take 4 holomorphic functions $ \{\phi(z)_\alpha, \psi(z)_\alpha\},~\alpha=1,2$ that satisfy a Dirac equation
$\partial_z \phi_\alpha=p\psi_\alpha,~\partial_{\bar{z}}\psi_\alpha=-p\phi_\alpha$
with real-valued $p(z,\bar{z})$. These define a conformal immersion into $\mathbb{R}^4$ with coordinates $X_a(z,\bar{z}), a=1,2,3,4$ that satisfy
$dX_1=\frac{i}{2}(\bar\phi_1\bar\phi_2+\psi_1\psi_2)dz+c.c.$
$dX_2=\frac{1}{2}(\bar\phi_1\bar\phi_2-\psi_1\psi_2)dz+c.c.$
$dX_3=-\frac{1}{2}(\bar\phi_1\psi_2+\bar\phi_1\psi_2)dz+c.c.$
$dX_4=\frac{i}{2}(\bar\phi_1\psi_2-\bar\phi_1\psi_2)dz+c.c.$
(for details on this "generalized Weierstrass representation", see Konopelchenko & Landolfi). Call this "immersed submanifold" $\Sigma$. Now I want to show these surfaces define a foliation, so I want to show the field of tangent planes is integrable. In these local coordinates I can write a vector field as
$ V=V^a \frac{\partial}{\partial X^a}=V^a(\frac{\partial z}{\partial X^a}\frac{\partial }{\partial z}+\frac{\partial \bar{z}}{\partial X^a}\frac{\partial}{\partial \bar{z}})$
It seems pretty obvious to me then that $[V,W]$ is a vector field on $\Sigma$ if $V,W$ are, and by Frobenius' Theorem this defines a 2-foliation. On the other hand, I know that any old surface in a smooth manifold does not necessarily define a 2-foliation and it didn't look like I did anything special except use coordinates which depend smoothly on the complex coordinate $z$. So am I right about this or did I do something strange?
 A: I believe you did something strange.
Any single independent pair of solutions $((\phi_1, \psi_1),(\phi_2,\psi_2))$ yields a single immersion $X: \mathbb{C} \to \mathbb{R}^4$ according to the paper you have cited. I'm guessing that this immersion is intended to be a covering map (otherwise, the image might not be a manifold) of a single differentiable surface $\Sigma$. Then you get a plane field $\tau$ (in fact, you get a 2-frame) as the span of the image of $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ under $X'$.
The problem is that $\tau$ is only defined on $\Sigma$, and not on all of $\mathbb{R}^4$. Frobenius's theorem would yield a 2-foliation only given a plane field on all of $\mathbb{R}^4$. As it is, we only have a plane field on $\Sigma$. So a 2-foliation of $\mathbb{R}^4$ including $\Sigma$ as a leaf exists only if (and if) we can extend $\tau$ to a plane field on all of $\mathbb{R}^4$.
EDIT: 
Let me put your comments this way. By the tubular neighborhood theorem, there is an embedding $i:(N\Sigma,\Sigma) \hookrightarrow (\mathbb{R}^4,\Sigma)$ of the normal bundle into $\mathbb{R}^4$ as a tubular, regular neighborhood $\mathcal{N}$ of $\Sigma$ in $\mathbb{R}^4$. We can get a plane field on $\mathcal{N}$ by pushing forward the obvious plane field $\tau$ on $\Sigma \times \mathbb{R}^2$, given by translations of the tangent plane field on $\Sigma$ in $\Sigma \times \mathbb{R}^2$. The pushforward $i_\ast(\tau)$ is a plane field on $\Sigma$ that restricts to the tangent plane field on $\Sigma$, and we get a 2-foliation on $\mathcal{N}$.
The problem is that $i:\mathcal{N} \hookrightarrow \mathbb{R}^4$ is not necessarily surjective. The points missed by $i$ don't get 2-plane elements assigned to them by $i$. To get a 2-foliation on all of $\mathbb{R}^4$, one must assign 2-plane elements to these missed points in a way that's consistent with $i_\ast(\tau)$. This may or may not be possible.
