# About isotopy of simple close curve

In the Primer mapping class group by farb Margalit. We have :

Proposition 1.10 Let $$\alpha$$ and $$\beta$$ be two essential simple closed curves in a surface $$S$$. Then $$\alpha$$ is isotopic to $$\beta$$ if and only if $$\alpha$$ is homotopic to $$\beta$$.

Proof. One direction is vacuous since an isotopy is a homotopy. So suppose that $$\alpha$$ is homotopic to $$\beta$$. We immediately have that $$i(\alpha, \beta)=0$$. By performing an isotopy of $$\alpha$$, we may assume that $$\alpha$$ is transverse to $$\beta$$. If $$\alpha$$ and $$\beta$$ are not disjoint, then by the bigon criterion they form a bigon. A bigon prescribes an isotopy that reduces intersection. Thus we may remove bigons one by one by isotopy until $$\alpha$$ and $$\beta$$ are disjoint.

In the remainder of the proof, we assume $$\chi(S)<0$$; the case $$\chi(S)=0$$ is similar, and the case $$\chi(S)>0$$ is easy. Choose lifts $$\widetilde{\alpha}$$ and $$\widetilde{\beta}$$ of $$\alpha$$ and $$\beta$$ that have the same endpoints in $$\partial \mathbb{H}^{2}$$. There is a hyperbolic isometry $$\phi$$ that leaves $$\widetilde{\alpha}$$ and $$\widetilde{\beta}$$ invariant and acts by translation on these lifts. As $$\widetilde{\alpha}$$ and $$\widetilde{\beta}$$ are disjoint, we may consider the region $$R$$ between them. The quotient $$R^{\prime}=$$ $$R /\langle\phi\rangle$$ is an annulus; indeed, it is a surface with two boundary components with an infinite cyclic fundamental group. A priori, the image $$R^{\prime \prime}$$ of $$R$$ in $$S$$ is a further quotient of $$R^{\prime}$$. However, since the covering map $$R^{\prime} \rightarrow R^{\prime \prime}$$ is single-sheeted on the boundary, it follows that $$R^{\prime} \approx R^{\prime \prime}$$. The annulus $$R^{\prime \prime}$$ between $$\alpha$$ and $$\beta$$ gives the desired isotopy .

I can't understand this proof for $$\chi(S)<0$$ . Is there any reference for more details for this proposition? Or simple proof ? I read paper by "bear" in the original proof but this is more difficult.

I think we consider $$\langle \phi\rangle$$ as discrate topological group acts on $$R$$ but why quotient $$R^{\prime}=$$ $$R /\langle\phi\rangle$$ is an annulus ? also I think $$\langle \phi\rangle$$ is isomorphic to infinite cyclic group $$\mathbb Z$$

Is this following shape true ? • To understand this proof you'll need at least theorem 1.2 from Farb Margalit, the classification of isometries of $\mathbb{H}^2$ (the isometry $\varphi$ the proof talks about is a hyperbolic isometry with axis the geodesic joining the endpoints of $\widetilde{\alpha}$ and $\widetilde{\beta}$), basic covering space theory and the obvious fact that if two curves are the boundary components of some annulus then they are isotopic. Jan 18, 2022 at 11:27
• @SaúlRodríguezMartín I read this theorem. But I can't understand why There is a hyperbolic isometry $\phi$ that leaves $\widetilde{\alpha}$ and $\widetilde{\beta}$ invariant and acts by translation on these lifts ? Why? Jan 18, 2022 at 11:33
• @SaúlRodríguezMartín .why The quotient $R^{\prime}=$ $R /\langle\phi\rangle$ is an annulus ? Jan 18, 2022 at 11:36
• @SaúlRodríguezMartín . What is $R /\langle\phi\rangle$ ? What is $\langle\phi\rangle$ ? Jan 18, 2022 at 11:44
• You could try my lecture notes here: dpmms.cam.ac.uk/~hjrw2/RS%20lectures.pdf . I wrote an updated version in the last few months -- I have a feeling one of the updates concerns this proof -- which I'll try to post in the next few days.
– HJRW
Jan 18, 2022 at 15:10

Suppose that $$S$$ is a closed connected surface with negative Euler characteristic. Suppose that $$H$$ is the universal cover of $$S$$. Equipping $$S$$ with a hyperbolic metric we find that $$H$$ is isometric to the hyperbolic plane.

Exercise: Every non-trivial deck transformation of $$H$$ over $$S$$ is hyperbolic: that is, has exactly two fixed points at infinity.

Suppose that $$\alpha$$ is a simple closed curve. Suppose that $$\alpha$$ is essential: in particular it does not bound a disk in $$S$$. We choose a point $$x_0$$ of $$\alpha$$ to be the basepoint of $$\pi_1(S, x_0)$$. Let $$y_0 \in H$$ be any preimage of $$x_0$$.

Let $$\phi : ([0, 1], \{0, 1\}) \to (S, x_0)$$ be a parameterisation of $$\alpha$$. Note that $$[\phi]$$ is an element of $$\pi_1(S, x_0)$$. We lift $$\phi$$ via path lifting to obtain a simple arc $$\phi_0 : ([0, 1], \{0, 1\}) \to (H, {y_0, y_1})$$. Note that path lifting defines the point $$y_1$$. There is a unique deck transformation that takes $$y_0$$ to $$y_1$$. In fact, this is the image of $$[\phi]$$ in the deck group. We will abuse notation and simply call this deck transformation $$[\phi]$$.

Exercise: $$y_1$$ is not equal to $$y_0$$. Thus the deck transformation $$[\phi]$$ is a hyperbolic transformation of $$H$$.

Let $$\langle \phi \rangle$$ be the subgroup of the deck group generated by $$[\phi]$$. Define $$\phi_k = [\phi]^k \cdot \phi_0$$. Let $$A = \cup \phi_k$$ be the union of the translates of $$\phi_0$$. We call $$A$$ an elevation of $$\alpha$$ (as it is a union of lifts giving a copy of the universal cover of $$\alpha$$).

Exercise: $$A$$ is a quasi-geodesic in $$H$$.

Thus $$A$$ has two endpoints in $$\partial H$$, the ideal boundary of $$H$$.

Exercise: Suppose that $$\beta$$ is a loop in $$S$$ that is (freely) homotopic to $$\alpha$$. Then all components of the preimage of $$\beta$$, in $$H$$, are quasi-geodesics, and exactly one (call it $$B$$) shares the ideal endpoints of $$A$$.

We now require an additional hypothesis. We must suppose that $$S$$ is orientable. If we do not, then it is possible that $$\alpha$$ is a core curve of a Möbius band: in this case $$\alpha$$ meets all curves in its free homotopy class. (You may want to think about how this interacts with the bigon criterion.)

Suppose that $$\beta$$ is a simple closed curve which is freely homotopic to, and disjoint from, $$\alpha$$. Recall that $$A$$ and $$B$$ are the chosen quasi-geodesic elevations of $$\alpha$$ and $$\beta$$. We deduce that $$A$$ and $$B$$ are disjoint in $$H$$ and share their ideal points in $$\partial H$$. So they cobound an (infinite) strip $$R$$ in $$H$$. Choose an arc $$c_0$$ that embeds in $$R$$ and connects $$x_0 \in A$$ to a point of $$B$$.

Exercise: We can choose $$c_0$$ so that it is disjoint from $$c_k = [\phi]^k \cdot c_0$$.

Finally, we deduce that $$c_0$$ and $$c_1 = [\phi] \cdot c_0$$ cobound a disk $$D$$ contained in $$R$$. Thus $$R / \langle \phi \rangle$$ is homeomorphic to the annulus $$D / (c_0 \sim c_1)$$

• So my shape is true ? Jan 18, 2022 at 18:07
• Neither of the pictures at the end of your post are an annulus. So, no? Jan 18, 2022 at 18:40
• Why the case $\chi(S)=0$ is similar? In this case we have euclidean surface but I don't know why we have similar way ? Jan 19, 2022 at 8:02
• Much of the argument is similar. However we will need to replace the hyperbolic geometry arguments with Euclidean ones. Jan 19, 2022 at 22:11
• is there any reference about your proof ? Jan 22, 2022 at 22:25