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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?

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1 Answer 1

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.

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I see, this is what I get for reading a physics book instead of a math book ;-) So my vectors $V,W$ are defined on an open subset $A\subset\Sigma$, but $A$ may not be an open subset of $M$. As a sanity check, this is because we are dealing with immersions right? If this was a topological embedding, then $A$ (now open in subspace topology) would be in the intersection $\Sigma\cap B$ for open set $B\subset M$. Then $V,W$ is defined on an open set of $M$ and we can use Frobenius. Any idea how to do your "cooking up" in the immersed case? –  cduston Jun 27 '11 at 17:59
    
To be clear: by $M$ you mean $\mathbb{R}^4$, right? Assuming I have understood you correctly, the problem is not immersed vs topological embedding but the fact yr vector fields are only defined on a $2$-dimensional submanifold of a $4$-dimensional space. If the submanifold is embedded then we can surely extend them to some open subset of $\mathbb{R}^4$ but would have no control over the Lie brackets of these extensions and so have no idea if Frobenius can be applied. –  Fran Burstall Jul 8 '11 at 10:59
    
Yes I mean $\mathbb{R}^4$, sorry about that. And it sounds like you have the problem stated correctly, but I just find it hard to believe (intuitively) that one cannot say $anything$ about extending vector fields to some open subset of $\mathbb{R}^4$. If we have an embedding and a metric, there should (locally) have some decomposition of the tangent space $T\mathbb{R}^4=T\Sigma \oplus TN$. Anyway, I'll soon mark the above as the answer for this question so as not to draw things out but I still feel there is more to be said here. –  cduston Jul 13 '11 at 1:43

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