I know that in the smooth category the following is true. There are at most countable many disjoint embedded moebius bands in euclidean 3-space. Is this also true in topological category?
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$\begingroup$ What exactly are you counting? Isotopy classes of single embeddings, or perhaps the number of disjoint embeddings? $\endgroup$– Ryan BudneyCommented Aug 6, 2011 at 17:04
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$\begingroup$ Disjont embeddings. $\endgroup$– MichalCommented Aug 6, 2011 at 17:09
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4$\begingroup$ How do you know this in the smooth category? $\endgroup$– Igor RivinCommented Aug 6, 2011 at 20:48
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1$\begingroup$ A few years ago I came across a paper by Grushin and Palamodov from 1962 called "On the maximal number of mutually disjoint, pairwise homeomorphic figures which can be packed in 3-space" (Uspekhi Mat. Nauk, 1962, Volume 17, Issue 3(105), Pages 163–168) where the case of Möbius strips is considered. Unfortunately the paper is in Russian and I know of no translations so I wasn't able to read it...and I am also not sure if this answers the OP's question or it's just the case he affirms to know already...It would be great if someone could tell me some details about what that paper says! $\endgroup$– godelianCommented Aug 7, 2011 at 1:49
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1 Answer
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There are at most countably many disjoint embeddings of homeomorphic images of a non-orientable hypersurface in $\mathbb R^k$. This is theorem 2 in "An uncountable family of disjoint spatial continua in Euclidean space" by V.K. Ionin and Yu.G. Nikonorov.
answered Aug 7, 2011 at 2:45
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$\begingroup$ Wow. Only countably many Moebius strips. Can you explain briefly which is the idea of the proof, or the heuristic reason why nonorientable things don't pack in a continuum? $\endgroup$– QfwfqCommented Aug 7, 2011 at 11:18
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5$\begingroup$ Well, as far as heuristics, non-orientable hypersurfaces are "thick", in the sense that two of them cannot be very close together, other wise one of them has to be the orientable double cover of the other. So if you consider their neighborhoods, an uncountable pigeonhole principle will give the desired contradiction. Compare to mathoverflow.net/questions/27244/… $\endgroup$ Commented Aug 7, 2011 at 12:19
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$\begingroup$ Actually, if you look at Young's paper (referenced in the question Gjergji references, the argument is a bit different from the pigeonhole principle. The basic idea there is that there is a region where the heads of the tacks foliate a ball, and so a point on the foot of the tack is separated from the head by a piece of another head, which contradicts either embeddedness or disjointness. As Gjergji says, nonorientable surfaces should be "thick" in some sense, which means that a similar argument should work. $\endgroup$ Commented Aug 7, 2011 at 14:05
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$\begingroup$ PS: Young's paper (which is one page) is available for free from the AMS (since it was published in BAMS). $\endgroup$ Commented Aug 7, 2011 at 14:10
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$\begingroup$ Reposting a link posted in a previous comment so that it appears in the "Linked" questions list: How many tacks fit in the plane? $\endgroup$ Commented Apr 5, 2023 at 7:15