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In this paper SOME EXTREMAL QUESTIONS FOR SIMPLICIAL COMPLEXES, the author discussed about minimal area of bounded by multiples of a curve. Say we have a (well-behaved) curve $\Gamma$, the solution to the minimal area problem is $S$. Now e consider $2\Gamma$, he solution to the minimal area problem is $S'$. But $area(S')<2area(S)$.

My question is what does $2\Gamma$ mean? Does it mean a similiar transformation in the coordinate multiplying by 2? If it is this case, then I cannot really imagine this is true.

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As @alesia points out $2\Gamma$ means to take the curve ``with multiplicity two" (this is usually understood to be in the context of currents or flat chains -- for an oriented curve you can think about it as tracing out the curve twice). The reason this can give a lower area for the minimizer is that the space of competitors for $2\Gamma$ contains two times all competitors for $\Gamma$ but may also contain new surfaces.

I'll try to illustrate this in (relatively) non-technical manner:

If $M$ is a Mobius band in $\mathbb{R}^3$ and $\Gamma=\partial M$, then it is the case that $2\Gamma=\partial \tilde{M}$ where $\tilde{M}$ is the orientation double cover of $M$. If it is the case that $M$ is the least area surface bounded by $\Gamma$ (among all orientable and non-orientable surfaces), then one should expect that the least area of orientable surface bounded by $\Gamma$, $N$, (which exists by appealing to geometric measure theory results) can satisfy $|N|>|M|$ (here $|N|$ and $|M|$ are the areas of $N$ and $M$, respectively). However, the least area orientable surface bounded by $2\Gamma$ should have area at most $2|M|$ (since $\tilde{M}$ is a valid competitor) and so one has $$|N'|\leq |\tilde{M}|= 2|M|<2|N|$$ where $N'$ is the least area orientable surface spanning $2\Gamma$.

EDIT: As pointed out in the comments the above ``example" doesn't work as $\tilde{M}$ can't have the claimed properties. The general idea is still correct and it seems can be made rigorous for a curve in $\mathbb{R}^4$.

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  • $\begingroup$ I apologise for commenting on an old thread, but I was curious about the relation of your remarks to White's paper called 'Regularity of area-minimizing hypersurfaces at boundaries with multiplicity'. As far as I understand, he shows that in codimension one an area-minimising current $T$ with $\partial T = \nu S$, where the multiplicity is $\nu \geq 2$ and where $S$ is a smooth, connected manifold can be decomposed into $T = T_1 + \cdots + T_\nu$ where $\partial T_i = S$. Could you clarify how it relates to your example with the Mobius band? ... $\endgroup$
    – Leo Moos
    Sep 3, 2021 at 12:14
  • $\begingroup$ In particular I thought that the least area of a current bounding $\nu S$ is exactly $\nu$ times the least area for a multiplicity one boundary. I think I am overlooking some issues related to the orientability, but my interpretation of White's results is that $\lvert N' \rvert = 2 \lvert N \rvert$. (By the way, from a technical point of view, in what space lies a non-orientable $M$ - is it OK here to consider it a mod two current?) $\endgroup$
    – Leo Moos
    Sep 3, 2021 at 12:20
  • $\begingroup$ @LeoMoos You may think of unoriented surfaces as mod 2 currents (see for instance this paper of Morgan ams.org/journals/tran/1984-283-01/S0002-9947-1984-0735418-6/…) where a Mobius strip minimizer is discussed. That being said looking at what I wrote, I can see it is not correct as the ``orientation double cover" is actually pushed forward to the 0 current so does not give a competitor. $\endgroup$
    – RBega2
    Sep 4, 2021 at 0:30
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I do not have access to the paper, but in this literature multiplying a curve by say $2$ roughly means taking the union of the curve with a very close translate of that curve. In reality, the translate actually coincides with the original curve, so we have "twice the same curve". This can be given a precise meaning in the formalism of currents.

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  • $\begingroup$ If they coincide how can the minimal surface bounded by that curve be different? $\endgroup$
    – Upc
    Apr 28, 2019 at 17:26
  • $\begingroup$ Without (eg topological) restriction, the minimal surface bounding the double of a curve actually has zero area (just take an infinitely thin ribbon). Also taking the same surface doesn't work, because the boundary would be the original curve, not its double (the boundary should go around the curve twice) $\endgroup$
    – alesia
    Apr 29, 2019 at 19:15
  • $\begingroup$ @alesia I just stumbled upon this question - sorry for making a comment so much later. However, I believe your remark (that the least area bounding $2 \Gamma$ is zero) is inaccurate. The thin ribbon you propose would give $\Gamma$ and its translate opposite orientations. As you take the limit and let the translate converge to the original $\Gamma$, this would give $0 [ \Gamma ]$, not $2 [ \Gamma ]$. $\endgroup$
    – Leo Moos
    Sep 3, 2021 at 12:32
  • $\begingroup$ @LeoMoos you're obviously right in the context of currents, although in the context of "plain" minimal surfaces (where orientation isn't taken into account) zero would be the answer. I really don't remember why I wrote this, but thanks for clarifying. $\endgroup$
    – alesia
    Sep 4, 2021 at 13:18

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