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How might one show that the set of connected 3D surfaces with infinite genus (up to homeomorphism) is countably infinite?

We could either use proof by contradiction or come up with a way to count them? What would be a relativelt simple way to show this?

Thanks.

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    $\begingroup$ What is a 3D surface? $\endgroup$
    – Igor Rivin
    Commented Sep 27, 2011 at 20:01
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    $\begingroup$ @IgorRivin: As in one in ${\mathbb R}^3$ $\endgroup$
    – James
    Commented Sep 27, 2011 at 20:11
  • $\begingroup$ You might find Benoit Kloeckner's answer at mathoverflow.net/questions/4155/… illuminating, which gives a classification of all non-compact surfaces. $\endgroup$
    – Jim Conant
    Commented Sep 27, 2011 at 21:13
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    $\begingroup$ By the way, there are uncountably many different embeddings of infinite genus surfaces in $\mathbb R^3$. Take a long tube, and put a line of tubes tied in knots along this tube. Letting these knots vary gives an uncountable family of surfaces, $\endgroup$
    – Jim Conant
    Commented Sep 27, 2011 at 21:18

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$\newcommand{\RR}{\mathbb R}$ $\newcommand{\HH}{\mathbb H}$ $\newcommand{\Ends}{\mathop{\rm Ends}}$

Let $F$ be an infinite genus surface, properly embedded in $\RR^3$. Then up to homeomorphism of surfaces (not ambient homeomorphisms) there are uncountably many such surfaces.

Proof. Note that $\RR^3$ is homeomorphic to $\HH^3$. So we may consider surfaces properly embedded in $\HH^3$ where there is more "room". Let $\HH^2$ be a totally geodesic plane in $\HH^3$. Let $T = T_3$ be a proper embedding of the regular $3$--valent tree into $\HH^2$. Take a small neighborhood of $T$ in $\HH^3$ to get a strange embedding of a three-ball $B$ into $\HH^3$.

Now, at every vertex of $T$ drill out a small tube from $B$ and take the boundary of the resulting infinite genus handlebody to get an infinite genus surface $F = F(T)$, properly embedded in $\HH^3$. Note that the topological space $\Ends(F)$, the ends of $F$, are an invariant of the homeomorphism type of $F$. In this example, $\Ends(F)$ is homeomorphic to the Cantor set.

By taking a subtree $T' \subset T$ we can obtain another surface $F' = F(T')$. As before, the $\Ends(F')$ is homeomorphic to the Gromov boundary of $T'$. Finally, since there are uncountably many pairwise non-homeomorphic compact subsets of the Cantor set, we find the required uncountable set of pairwise non-homeomorphic embedded surfaces. QED

This answer was partly prompted by Agol's answer to a previous question.

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  • $\begingroup$ In the construction of the surfaces that you mentioned, instead of a 3-valent tree we could as well use an $n$-valent tree (with $n\geq 3$). Also instead of adding one handle at each vertex we could add any positive number g of handles. Would the surfaces so obtained be homeomorphic to each other independent of choice of g and n ? $\endgroup$
    – user10001
    Commented Jul 13, 2014 at 11:17
  • $\begingroup$ I believe that, as long as $n \geq 3$ and $g \geq 1$, all of surfaces you mention are homeomorphic. The homeomorphisms are pretty crazy -- you have to "borrow" genus from increasingly distant vertices. $\endgroup$
    – Sam Nead
    Commented Jul 13, 2014 at 16:07

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