Counting submanifolds of the plane - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T01:11:56Z http://mathoverflow.net/feeds/question/25009 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25009/counting-submanifolds-of-the-plane Counting submanifolds of the plane Sam Nead 2010-05-17T13:41:58Z 2010-07-14T18:44:42Z <p>After thinking about this <a href="http://mathoverflow.net/questions/24932/counting-connected-manifolds/24938#24938" rel="nofollow">question</a> and reading this <a href="http://mathoverflow.net/questions/4155/classification-problem-for-non-compact-manifolds/11259#11259" rel="nofollow">one</a> I am led to ask for an uncountable collection of homeomorphism types of boundaryless connected path-connected submanifolds of the plane. </p> <p>My guess is that is suffices to consider complements of Cantor sets. However, I do not know how to distinguish ends (up to homeomorphism) sufficiently well to ensure that this works. Are there other, easier, invariants?</p> http://mathoverflow.net/questions/25009/counting-submanifolds-of-the-plane/25039#25039 Answer by Agol for Counting submanifolds of the plane Agol 2010-05-17T17:44:05Z 2010-07-14T18:44:42Z <p>See a <a href="http://www.ams.org/mathscinet-getitem?mr=143186" rel="nofollow">theorem of Richards</a>, which implies that homeomorphism types of planar surfaces are in 1-1 correspondence with homeo. types of compact subsets of the Cantor set. I think there should be uncountably many homeo. types of totally disconnected compactums, but I don't know a reference or an argument off the top of my head. I think this should be related to the ordinalities of the accumulation points, but I'm not sure which ordinals can occur. </p> <p>Addendum: Googling, I found references to a result of Markiewicz-Sierpinski classifying countable compact metric spaces up to homeomorphism by their <a href="http://en.wikipedia.org/wiki/Derived_set_%28mathematics%29#Cantor.E2.80.93Bendixson_rank" rel="nofollow">Cantor-Bendixson rank</a> (see <a href="http://arxiv.org/abs/0710.1495" rel="nofollow">section 3 of this paper</a> for a statement). The CB-rank must be a countable <a href="http://en.wikipedia.org/wiki/Ordinal_number" rel="nofollow">ordinal</a> $\zeta$, and the space is homeomorphic to the ordinal $\omega^\zeta\cdot n+1$ with the order topology for some $n\in \mathbb{N}$. These may all be realized as compact subsets of the line. This gives uncountably many non-homeomorphic compacta, which by Richards' theorem implies that there are uncountably many planar surfaces. </p>