Let $S$ be a compact orientable 2-surface with $\chi(X)\leq -3$. Y.Minsky and H.Masur proved that the curve complex of $X$ is $\delta$-hyperbolic and infinite. E.Klarreich (see also U.Hamenstadt) proved the Gromov boundary of the curve complex of $S$ is bijective to the collection of ending laminations. Denote the collection of ending laminations by $B$.

Note: ending lamination implies that its complement in $S$ is a collection of (once-punctured) ideal polygons.

My question is: Given a number $N$, is there possible that there is a collection of essential simple closed curves $Y=\{c_{\eta} \mid \eta\in B\}$ such that $d_{\mathcal {C}(S)}(c_{\eta}, c_{\zeta})\geq N$, for any $\eta, \zeta \in B$ and $\eta \neq \zeta$?

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    $\begingroup$ What do you mean by "$Y=c_\eta, \eta\in B$"? What is $c_eta$? Do you mean that $c_\eta$ is a sequence of scc's $c_{\eta,n}$ representing the ideal point $\eta$? Then the answer is obviously yes: $$ \lim_{n\to\infty} d(c_{\eta,n}, c_{\zeta,n})=\infty, $$ and it holds for any $\delta$-hyperbolic geodesic space, simply by the definition of Gromov boundary. $\endgroup$ – Misha Nov 4 '12 at 14:06
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    $\begingroup$ Actually, you do not even need the space to be geodesic for this to hold, although the curve complex is geodesic. $\endgroup$ – Misha Nov 4 '12 at 14:52
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    $\begingroup$ To correct your reference, the theorem equating the Gromov boundary of the curve complex to the space of ``ending laminations'' is due not to Hamenstadt but to Klarreich www.ericaklarreich.com/page15.html $\endgroup$ – Lee Mosher Nov 4 '12 at 17:10
  • $\begingroup$ @Misha, $Y$ is defined to be a collection of essential closed curves. I want to write it as a collection, but I don't know how to do it. And the number $N$ depends only on the surface $S$. $\endgroup$ – yanqing Nov 4 '12 at 23:34
  • $\begingroup$ @Lee Mosher, you are right. I just correct it. $\endgroup$ – yanqing Nov 4 '12 at 23:37

Are you trying to index a countable collection of curves by the (uncountable!) space of ending laminations?

  • $\begingroup$ Who knows what OP means by $c_\eta$: Could be a single vertex (then, of course, the map is not going to be 1-1), could be a sequence, could be something completely different... $\endgroup$ – Misha Nov 5 '12 at 23:15
  • $\begingroup$ $c_{\eta}$ is a single vertex. $\endgroup$ – yanqing Nov 6 '12 at 0:02
  • $\begingroup$ Then anonymous answered your question, since you cannot have an injective map from continuum to a countable set. $\endgroup$ – Misha Nov 6 '12 at 0:09
  • $\begingroup$ @staylor, I am not sure that the vertices of the curve complex is countable. $\endgroup$ – yanqing Nov 10 '12 at 0:10
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    $\begingroup$ The fundamental group of your surface is countable (it is finitely generated) and homotopy classes of closed curves correspond to conjugacy classes in this group. The vertices of the curve complex are homotopy classes of essential simple closed curves. $\endgroup$ – staylor Nov 10 '12 at 18:00

Just to summarize the above: the answer is "no" because the curve complex has countably many vertices while the boundary is uncountable. To give an easier example of this sort of thing, consider the regular four-valent tree, also known as the Cayley graph of the rank two free group. The tree is countable while its Gromov boundary is a Cantor set, and so uncountable.


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