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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?


  • The question is inspired by this; you will find there a construction 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 $f_1|_ {\partial D^2}\equiv f_2|_ {\partial D^2}$.

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I had almost the same solution, but I was on the train. One more question - is it possible to prove that there are parallel tangent planes at inner points of disks? I do not know. –  Petya Mar 8 '10 at 22:18
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2 Answers

up vote 12 down vote accepted

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.

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Thanks, I was so sure such monsters do live so did not even try prove opposite :) –  Anton Petrunin Mar 8 '10 at 18:03
Anton, if you are hunting for monsters, then I can tell you that your question is open for cylinders (regular homotopy rel boundary). In that case a topological obstruction is zero. The existence of such monsters might be a corollary of Novikov-Khovanskii theory of convex-concave sets. I'll try to check it. –  Petya Mar 9 '10 at 11:57
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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 (http://en.wikipedia.org/wiki/Borsuk–Ulam_theorem).

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Why is it impossible for $\Sigma_1\cup \Sigma_2$ to have a global (non-vanishing) vector field? –  Paul Mar 8 '10 at 19:35
Since $\Sigma_1 \cup \Sigma_2$ is a $2$-sphere! (Or better, since the identification of $\Sigma_1$ with $\Sigma_2$ along their boundary is a 2-sphere.) d –  damiano Mar 8 '10 at 20:04
I have accepted the answer of Steven since it was few minutes earlier. (I would accept both if I could.) –  Anton Petrunin Mar 8 '10 at 21:33
@daimiano Yes, but $\Sigma_1\cup \Sigma_2$ is not a smoothly embedded 2-sphere (if it were then the tangent planes along the boundary would agree already). E.g. if $\Sigma_1=\Sigma_2$ there is a nowhere zero vector field on the union. I must be missing the point. The extra fact in Steven's answer is that his vf is tangent to the boundary circle. Perhaps this can be arranged in your argument using the cross product. –  Paul Mar 8 '10 at 23:19
@paul: i see your point now. i think that, as it stands, my argument requires smoothness (or at least $C^1$) up to the boundary. it might be possible to get rid of this by small perturbations, but the accepted proof is much simpler that i do not feel inclined to pursue this course of action! Btw, I had in mind an "antipodal" identification, not a direct one, if this helps. In the case $\Sigma_1=\Sigma_2$ my "proof" should induce a "proof" of the Brouwer fixed point theorem. –  damiano Mar 9 '10 at 9:42
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