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, arXiv:math\9804144v3). 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?



Conformally immersed Riemann surfaces and foliations
Um, $V,W$ are vector fields on the *surface* but not on $\mathbb{R}^4$. You need a distribution of $2$-planes on (an open subset of) $\mathbb{R}^4$ to use Frobenius to cook up a foliation.


