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Jun 24, 2021 at 17:27 history edited RBega2 CC BY-SA 4.0
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Jun 24, 2021 at 11:10 comment added mlk @LeoMoos Yes, you are right, I did not read that comment.
Jun 24, 2021 at 10:56 comment added Leo Moos @mlk Doesn't Otis's comment above point to the fact that the comparison argument is inapplicable for currents? As far as I understand when $T$ is the union of countably many parallel planes, then the comparison gives a bound like $R \mathcal{H}^n(\partial B_R)$ because there are roughly $R$ planes intersecting $B_R$.
Jun 24, 2021 at 9:48 comment added mlk Personally in the case of currents I think one would simply prefer slicing because it is the more natural operation. In particular you'd get some continuity properties for free. That being said though, for currents and in fact for any orientable surfaces, you'd need some more work, because while by standard arguments there is always a minimal current on the sphere with the prescribed boundary, it may have multiplicity larger than 1 in some places (e.g. if $\tau$ consists of several identically oriented circles on the same hemisphere).
Jun 24, 2021 at 9:45 comment added Leo Moos I think you can disregard my previous comment, at least when the dimension is large enough, say $n \geq 8$ that the singular set would be larger than a set of isolated points. I appreciate your explanation regarding the slicing argument.
Jun 24, 2021 at 0:20 comment added RBega2 You're probably right, though I guess one would need to be a little worried about the argument being circular (i.e., the area bound is almost certainly used implicitly in the regularity theory). Slicing (of currents) is useful as then you would obtain a slice that is a closed current in the boundary of the ball, so the region $\Omega$ could also be a current whose boundary was a slice
Jun 23, 2021 at 23:33 comment added Leo Moos I'm just curious: why would slicing be required when working with currents? The singular set has codimension seven, so $\partial B_R \cap T$ (or $\partial B_R \cap \mathrm{spt} \lVert T \rVert$ to be precise) would still be regular for generic $R$, no? Mind you I've never seen slicing 'in the wild', so I'm not positive how it's usually applied.
Jun 23, 2021 at 23:20 vote accept Leo Moos
Jun 23, 2021 at 21:37 history answered RBega2 CC BY-SA 4.0