Two discs with no parallel tangent planes 
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?

Comments:


*

*The question is inspired by this; 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 $f_1|_ {\partial D^2}\equiv f_2|_ {\partial D^2}$. 
 A: If I understood your question correctly, I think that the answer is no.  In fact, even more is true: if you choose any identification $\varphi \colon \Sigma_1 \to \Sigma_2$ in such a way that the boundary are compatibly identified, then for any embedding of $\Sigma_1$ and $\Sigma_2$ there is a point $p$ of $\Sigma_1$ such that the tangent plane to $\Sigma_1$ is the same as the tangent plane to $\Sigma_2$ at $\varphi(p)$.  To see this, let $\mathbb{R}P^2$ denote the real projective plane, the space parameterizing linear subspaces of dimension one in $\mathbb{R}^3$.  Suppose by contradiction that you found embeddings with the mentioned property.  Let $\gamma \colon \Sigma_1 \cup \Sigma_2 \to \mathbb{R}P^2$ be the function defined by sending $p$ to the linear subspace spanned by $N_p \wedge N_{\varphi(p)}$, where $N_p$ is the normal direction to the image of $p$ and similarly for $N_{\varphi(p)}$, and the wedge product is the usual wedge product in $\mathbb{R}^3$.  The function $\varphi$ is indeed a function, since never the normal directions at $p$ and $\varphi(p)$ are parallel.  But observe that $\gamma(p)$ is always a vector contained in the tangent plane at $p$.  Thus, $\gamma$ determines a global vector field on the image of $\Sigma_1 \cup \Sigma_2$ that is never zero.  This is obviously impossible!
Note that the above argument is essentially the one used in the Borsuk–Ulam Theorem.
A: 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.
