In this paper [SOME EXTREMAL QUESTIONS FOR SIMPLICIAL COMPLEXES](https://link.springer.com/content/pdf/10.1007/BF02851262.pdf), 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.



Least area bounded by multiple of curves
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 $\mathbb{R}^4$.


