# Every closed $C^1$ curve in $\mathbb R^3 \setminus \{ 0 \}$ is the boundary of some $C^1$ 2-surface $\Sigma \subset \mathbb R^3 \setminus \{ 0 \}$

How can I prove it?

This problem looks similar to Plateau's problem - but it is much more specific. I believe there exists some elementary proof.

(Proving this will help me apply Stokes' theorem to another problem)

EDIT: I have a proof that works in some cases (I hope). Assume $\Gamma$ is a such a curve:

• Calculate the center of mass for the curve.
• Translate the curve by its center of mass.
• Define $\Sigma'=\cup_{0 \leq \lambda \leq 1} \lambda \Gamma$, where $\lambda \Gamma=\{ \lambda p:p \in \Gamma\}$
• $\Sigma$ will be $\Sigma'$ translated backwards by the center of mass.
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What smoothness requirement are you placing on $\Sigma$? –  Alex Becker Mar 23 '13 at 21:58
$C^1$ as well, sorry. I also added that $\Gamma$ should be closed. –  user1337 Mar 23 '13 at 21:59
You can only expect an immersed surface, not an embedded surface. To see this, consider e.g. the case where the curve is a trefoil knot in $\mathbb{R}^3$. –  Hans Engler Mar 23 '13 at 23:11

Since your goal is to apply Stokes' theorem, you need an orientable surface. The orientability makes the problem tricky, which is why it took until 1930 to sort it out. Such a surface indeed exists for any smooth $\Gamma$; it is called a Seifert surface of $\Gamma$ and can be obtained using Seifert's algorithm. This page has a schematic drawing of a Seifert surface of the trefoil knot; more detailed images can be found with a Google search; e.g., Visualization of Seifert surfaces.
Your candidate for $\Sigma'$ is not a smooth surface unless $\Gamma$ is a planar curve (look at what happens at the center of mass). It may have self-intersections issues too.
What do you mean by smooth? $C^1$ or $C^\infty$? –  user1337 Mar 24 '13 at 8:47