0
$\begingroup$

In the paper "Geometry of the complex of curves II: Hierarchical structure" (Paper) there is a construction of curve complex for an Annular subdomain (2.4). The construction depends on the domain itself but the definition of domain consist of isotopy classes of subsurfaces. So I have the following doubts.

1) Why the construction is unique upto isotopy?

2) Curves are considered upto isotopy, then what do they mean by 'The' core curve?

3) The vertices of curve complex are isotopy classes of curves. Then why the function $\pi_Y$ defined in the next paragraph is well defined?

PS: In para 2.3 they said that "to make discussion clear one might fix a complete hyperbolic metric on $S$." So does this metric assumption is necessary for those constructions or we can do them topologically.

$\endgroup$

1 Answer 1

4
$\begingroup$

1) Why the construction is unique upto isotopy?

Recall that $S$ is the surface and $Y$ is the annular subsurface. Masur and Minsy take an annular cover $\tilde{Y}$ and work there. The cover $\tilde{Y}$ depends only on the free homotopy class of any core curve $\alpha \subset Y$.

2) Curves are considered upto isotopy, then what do they mean by 'The' core curve?

They define "curve" in the first sentence of page 12. Also, note that they write $\alpha \in C_0(S)$, so they mean the isotopy class.

It is standard practice to use the same notation for curves and their isotopy classes. Tracking the difference between $\alpha$ and $[\alpha]$ is often not necessary. I grant that this abuse of notation can cause confusion the first time you see it.

3) The vertices of curve complex are isotopy classes of curves. Then why the function $\pi_Y$ defined in the next paragraph is well defined?

The definition of the "curve complex" $C(Y)$ is altered when $Y$ is an annulus. Instead of using isotopy classes of essential curves they use isotopy classes rel endpoints of essential arcs in the Gromov closure of $\tilde{Y}$. The classes downstairs (in $S$) lift to the correct classes upstairs (in $\hat{Y}$, the Gromov closure).

$\endgroup$
7
  • $\begingroup$ Thank you very much for the clarification. But I am still confused about the third part. The Gromov clouser requires the cover to have a negatively curved metric right! So are we considering $\bar{Y}$ as a quotient of $\mathbb{H}^2?$ because then it is same as considering a hyperbolic metric in $S$. $\endgroup$
    – Cusp
    May 1, 2014 at 10:42
  • $\begingroup$ You can do things this way. However, taking the Gromov closure only requires that the metric space in question be $\delta$-hyperbolic. It is an exercise to check that for any metric $\rho$ on $S$ (with $\chi(S) < 0$) the annular cover is $\delta$-hyperbolic. (However, it is true that $\delta$ depends $\rho$.) $\endgroup$
    – Sam Nead
    May 1, 2014 at 10:47
  • $\begingroup$ Again thanks. Actually I was trying to understand this from a complete topological background, so I was trying to avoid any use of geometry. $\endgroup$
    – Cusp
    May 1, 2014 at 10:53
  • $\begingroup$ Perhaps I should have said "It is a non-trivial exercise..." $\endgroup$
    – Sam Nead
    May 1, 2014 at 10:53
  • $\begingroup$ Its okay. I knew that result. $\endgroup$
    – Cusp
    May 1, 2014 at 10:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.