Let $K$ be a knot smooth knot in a 3-manifold $M$ and fix a metric on $M$. Let $F$ be a orientable surface of genus $g$ with one boundary component. Then we can consider the family of all maps $\mathscr{F} = \{ \phi: (F, \partial F) \to (M,K) : \phi \text{ is an embedding} \}$. By pulling back the metric we can talk about $\text{Area}(\phi)$ so we can ask if some element of $\mathscr{F}$ achieves the minimum area amongst all elements of $\mathscr{F}$. I suppose that I am mainly interested in knots in $S^3$ with the usual metric, if that simplifies things.

(1) Is this minimum achieved?

I have seen this paper where Proposition 1 might address my question - however, they allow for piecewise smooth maps. Is there any reason why the area minimizing surfaces would need to be smooth? This seems intuitively clear, but I am not familiar with how to prove such things.

(2) If the minimum is achieved, is it achieved by a genus minimizing surface? (I.E. the genus of $F$ is the genus of the knot).


In question (1), if you allow $g$ to vary, then this is answered positively by Hardt and Simon (see also).

The answer to question (2) is no. Almgren and Thurston construct unknots which do not bound an embedded disk in their convex hull. If $\mathscr{F}$ (fixing $g=0$) contained a minimal area member, then it would have to be a minimal surface. However, a minimal surface bounding $K$ must lie in the convex hull of $K$, a contradiction.

  • $\begingroup$ It looks like two answers contradict to each other completely. I guess that Danny Ruberman got it wrong? $\endgroup$ – aglearner Dec 14 '18 at 23:40
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    $\begingroup$ @aglearner: Danny's question refers to a minimal surface with respect to a metric on the knot complement in which the boundary (a torus) is convex. So in that case, his answer is correct. However, still in that case, I'm not sure that (2) is true, depending on the metric. From the statement of your question, I assumed you wanted a minimal surface bounding the knot with respect to the ambient metric. $\endgroup$ – Ian Agol Dec 14 '18 at 23:45
  • $\begingroup$ Thank you for explaining! (that was not my question, but that of user101010, so I could not understand that Danny treats only a very special subcase of the question). $\endgroup$ – aglearner Dec 15 '18 at 0:01
  • $\begingroup$ My answer wasn't very clear; I was addressing the question raised in the original post about Edmonds' pace. As others have remarked this does not seem to be what was intended by the OP. As @IanAgol says, Edmonds uses a metric on the complement of the knot (with convex boundary), not the ambient metric. $\endgroup$ – Danny Ruberman Dec 15 '18 at 15:36
  • $\begingroup$ @IanAgol Thanks Ian, this helps me piece things together although it will take me some time to look through these. Is there any reason to think that the answer to (1) for a fixed $g$ would be no? $\endgroup$ – user101010 Dec 15 '18 at 18:09

Yes, the surface is smooth, and you can get one of minimal genus. This follows from the references in the paper of Edmonds that you cite. The paper of Freedman-Hass-Scott shows that the least area surface in the homotopy class of a minimal genus Seifert surface is actually smooth. It is a useful observation that this minimization also works in the larger class of piecewise smooth surfaces. This comes up when (as in Edmonds's paper) you are doing cut/paste arguments with a pair of surfaces.

  • $\begingroup$ Thanks - when looking at the Edmonds paper, my eyes glazed over at the part where he assumes the boundary is convex. I guess that really is a significant assumption that I should have highlighted. $\endgroup$ – user101010 Dec 15 '18 at 18:07

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