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> Is it possible to construct smooth embedded of 2-discs $\Sigma\_1$ and $\Sigma\_2$ in $\mathbb R^3$ with the same boundary curve such that there is no pair of points $p\_1\in \Sigma\_1$ and $p\_2\in \Sigma\_2$ with parallel tangent planes?
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**Comments:**


* The question is inspired by [this](https://mathoverflow.net/questions/16335); you will find there a construction of three such discs with no triples of points.
* More formally, *"the same boundary curve"* means that $\Sigma\_1=f\_1(D^2)$ and $\Sigma\_2=f\_2(D^2)$ for some smooth embedding $f\_1$ and $f\_2$ of the unit disc $D^2$ 
such that $f\_1|\_ {\partial D^2}\equiv f\_2|\_ {\partial D^2}$.



Two discs with no parallel tangent planes
Suppose two such disks Σ1 and Σ2 exist, and pull back TΣ2 to Σ1 by some homeomorphism. Viewed as a subbundle of TR3|Σ1, this plane bundle intersects TΣ1 in a line bundle L over Σ1 since no two tangent planes are parallel. Furthermore, L|∂Σ1 is exactly the bundle of lines tangent to ∂Σ1.


Since a line bundle over a disk is trivial, we can take a nonzero section of L, and thus we get a nonvanishing vector field on the disk Σ1 which is tangent to the boundary at every point of ∂Σ1. But then by doubling we can construct a nonvanishing vector field on S2, and this is impossible.


