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I am reading the third chapter on "Dehn Twists" from "A Primer on Mapping class group" by Benson Farb and Dan Margalit. I am stuck on the definition of Dehn Twist. The book defines the Dehn Twist as follows. First, define the twist map on the annulus as follows $\mathbb{S}^{1}\times [0,1] \to \mathbb{S}^{1} \times [0,1]$ as $(\theta, t) \mapsto (\theta + 2\pi t, t)$. It is clear that this defines an orientation preserving homeomorphims preserving the boundary pointwise. We define the Dehn Twist as follows: Let $S$ be an arbitrary oriented surface and let $\alpha$ be a simple closed curve in $S$. Let $N$ be a regular neighborhood of $\alpha$ and choose an orientation preserving homeomorphism $\phi: A \to N$. We obtain an orientation preserving homeomorphism $T_{\alpha}: S \to S$, called the Dehn Twist about $\alpha$ as follows: $T_{\alpha}(x) = \phi \circ T \circ \phi^{-1}(x)$ when $x \in N$ and $T_{\alpha}(x) = x$ when $x\in S \setminus N$.

My Question is that: Why is the Dehn Twist an well-defined representative of the mapping class group. Meaning,

(1) Why is the (isotopy class of) Dehn Twist $T_{\alpha}$ invariant under the choice of the neighborhood $N$?

(2) If $\alpha$ and $\beta$ are two isotopic simple closed curves on the surface when should the induced Dehn Twists be isotopic/homotopic?

Jakobian
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Dwaipayan Sharma
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    You would need to show that any two such maps $\phi_0, \phi_1 : A \to S$ which arise in this way from the same $\alpha$ are (ambient) isotopic. With such an isotopy in hand, it becomes clear that the Dehn twist homeomorphisms arising from $\phi_0, \phi_1$ are themselves isotopic. A similar argument works in the smooth case and will likely be even easier. Unfortunately I don’t know the details for how to show $\phi_0, \phi_1$ are isotopic. – Mark Saving Feb 19 '25 at 04:49
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    By the way, I'll point out that the question that you really ask in (1) and (2) is not why the Dehn Twist is a well-defined element of the mapping class group, but instead, why a Dehn Twist is a well-defined representative of its mapping class. – Lee Mosher Feb 19 '25 at 23:29
  • Yeah that was what I meant. I have edited it now. Thanks. – Dwaipayan Sharma Feb 20 '25 at 01:29

1 Answers1

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Here's an outline of the proof in a few steps.

Let $N_\alpha,N_\beta$ be the closed regular neighborhoods of $\alpha$ and $\beta$ on which the twist maps are defined. We also have chosen orientation preserving homeomorphisms $\phi_\alpha : A \to N_\alpha$ and $\phi_\beta : A \to N_\beta$ with $A = \mathbb S^1 \times [0,1]$.

First, we can assume that $N_\alpha=N_\beta$, for the following reason. Knowing that $\alpha$ and $\beta$ are isotopic, it follows that their regular neighborhoods $N_\alpha$ and $N_\beta$ are "ambient" isotopic, which means that there exists a continuous function $h : S \times [0,1] \to S$ such that each map $h_t(x)=h(x,t)$ is a homeomorphism, $h(x,0)=x$, and $h(N_\alpha,1)=N_\beta$. The formula $$h_t^{-1} \circ T_\beta \circ h_t $$ then defines an isotopy from $$T_\beta = h_0^{-1} T_\beta h_0 $$ to a Dehn twist around $\alpha$ denoted $$T'_\alpha = h_1^{-1} T_\beta h_1 $$ and we conclude that $T'_\alpha$ has the exact same twist neighborhood $N_\alpha$ as $T_\alpha$. In summary: we can use an ambient isotopy from $\alpha$ to $\beta$ to conjugate $T_\beta$ to $T'_\alpha$.

Furthermore, although we don't yet have $\phi_\alpha=\phi_\beta$, and so $T_\alpha$ and $T'_\alpha$ need not be equal on $N_\alpha$, what we do have already is that $T_\alpha$ and $T'_\alpha$ both restrict to the identity map on $S - \text{interior}(N_\alpha)$, and so both also restrict to the identity map on the boundary of the annulus $N_\alpha$. This is clear for $T_\alpha$, and for $T'_\alpha$ it follows from the formula used above to define $T'_\alpha$, together with the fact that $T_\beta$ restricts to the identity map on $S - \text{interior}(N_\beta)$.

Using $\phi_\alpha : A = \mathbb S^1 \times [0,1] \to N_\alpha$ consider the topological arc $J = \phi_\alpha(p \times [0,1]) \subset N_\alpha$ (pick $p \in S^1$ arbitrarily).

Next, we can assume that $T_\alpha(J)=T'_\alpha(J)$, for the following reasons. We know that arcs $J_\alpha=T_\alpha(J)$ and $J'_\alpha=T'_\alpha(J)$ are isotopic rel endpoints in the annulus $N_\alpha$; this is easiest to see in the universal cover $\mathbb R \times [0,1]$ of $A \approx N_\alpha$, where $J$ lifts to $\tilde p \times [0,1]$, and where $T_\alpha$ and $T'_\alpha$ have lifts each of which is the identity on $\mathbb R \times 0$ and which shifts one unit to the right on $\mathbb R \times 1$. It follows that $J_\alpha$ and $J'_\alpha$ are ambient isotopic in $N_\alpha$, using an ambient isotopy which is the stationary on the boundary of $N_\alpha$. Again, by conjugating using this ambient isotopy, we arrange that $T_\alpha(J)=T_\beta(J)$, and everything is still the identity outside of $\text{interior}(N_\alpha)$.

Next, we can assume that $T_\alpha \mid J = T'_\alpha \mid J$: we use here the fact that two homeomorphisms between two arcs which have the same endpoint values are isotopic; that isotopy can then be extended to an ambient isotopy on the surface which is still the identity outside of $\text{interior}(N_\alpha)$, and again we conjugate using this ambient isotopy.

Finally, represent the annulus $N_\alpha$ as the image of the square $[0,1] \times [0,1]$ by gluing the top to the bottom of the square. The map $T_\alpha^{-1} T'_\alpha$ pulls back to a homeomorphism of this square which is the identity on the boundary, and it remains to show that this homeomorphism is isotopic rel boundary to the identity map on the square. This is just the Alexander trick.

Lee Mosher
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  • I don't understand the 6th paragraph. "Next we can assume $T_{\alpha}(J) = T_{\alpha}^{'}(J)$. Since $T_{\alpha}$ and $T_{\alpha}^{'}$ are isotopic...." Aren't we proving $T_{\alpha}$ and $T_{\alpha}^{'}$ are isotopic? – Dwaipayan Sharma Feb 20 '25 at 02:30
  • The last paragraph is clear to me. But I don't understand why do we need to build through the topological arc argument that is, why do we need $T_{\alpha}|{J} = T{\alpha}^{'}|_{J}$. Sorry for the silly question. – Dwaipayan Sharma Feb 20 '25 at 08:29
  • Yeah, that 6th paragraph did not come out well. I rewrote it. – Lee Mosher Feb 20 '25 at 13:36
  • And the reason we need $T_\alpha \mid J = T'\alpha \mid J$ is to get the consequence that when we pull the map $T\alpha^{-1} T'_\alpha$ back to a self-homeomorphism of the square, what we get restricts to the identity map on the boundary of the square. – Lee Mosher Feb 20 '25 at 13:37
  • @LeeMosher But by your definition, both $T_{\alpha}$ and $T_{\alpha}^{'}$ are fixing the boundary of $N_{\alpha}$ pointwise. Doesn't it entail we would always get the pull back homeomorphism between the squares to be fixing the boundary pointwise? –  Feb 21 '25 at 05:30
  • Oh I think I am making a mistake, the square has 4 boundary components while the annulus has 2, the penultimate paragraph is necessary to imply the pull back homeomorphism fixes all of the four boundary components pointwise. –  Feb 21 '25 at 05:33
  • Also I cannot see how you are concluding $T_{\alpha}^{-1}T_{\alpha}^{'}$ are fixes the boundary pointwise. –  Feb 21 '25 at 06:04
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    Once you know that $T_\alpha$ and $T'\alpha$ each fix every point on the boundary of $A$, it follows that $T{\alpha}^{-1} \circ T'\alpha$ also fixes every point on the boundary of $A$. And once you know that $T\alpha \mid J = T'\alpha \mid J$, it follows that $T\alpha \circ T'\alpha$ fixes every point on $A$. And then, since the boundary of the square maps to the boundary of $A$ union $J$, when you pull $T\alpha^{-1} \circ T'_\alpha$ back to the square the resulting homeomorphism of the square fixes every point on the boundary of the square. – Lee Mosher Feb 21 '25 at 13:00
  • Sorry but I still don't see how $T_{\alpha}|{J} = T^{'}{\alpha} |{J}$. Can you please elaborate on this fact? We want to conclude that $T{\alpha}^{-1}T_{\alpha}^{'} = id$ on J. How to do that?? –  Feb 24 '25 at 09:28
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    We are applying a theorem about $[0,1]$, namely, for every homeomorphism $g : [0,1] \to [0,1]$, if $g(0)=0$ and $g(1)=1$ then $g$ is isotopic to the identity. This is the 1-dimensional version of the Alexander trick. – Lee Mosher Feb 24 '25 at 14:08
  • Yes but that only entails that $T_{\alpha}|{J}$ and $T{\alpha}^{'}|_{J}$ are isotopic, how can we assume they are equal? I don't understand how isotopy is equality. –  Feb 24 '25 at 16:21
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    @LeeMosher In your previous paragraph you have established, $T_{\alpha}(J)$ is isotopic (via an ambient isotopy of $N_{\alpha})$ to $T_{\alpha}^{'}(J)$. Call that ambient isotopy to $k: N_{\alpha} \times I\to N_{\alpha}$ such that $k_{1}(T_{\alpha}(J)) = T_{\alpha}^{'}(J)$ . That is, $T_{\alpha}^{-1}\circ k_{1}^{-1}\circ T_{\alpha}^{'}$ is identity on J. Then its pull back on the square will be isotopic to identity by the Alexander trick. Due to similar reasons $k_{1}T_{\alpha}$ is isotopic to $T_{\alpha}$. Hence $T_{\alpha}^{'}$ is isotopic to $T_{\alpha}$. Does this work? –  Feb 24 '25 at 16:45
  • $k: N(\alpha) \times I \to N(\alpha)$ such that $k_{t}(x) = k(x,t)$. I think in addition to that we would need to choose the isotopy $k$ preserves the boundary of $N_{\alpha}$ pointwise. –  Feb 24 '25 at 16:49
  • You might have overlooked something in what I wrote. When reducing to the case $N_\alpha=N_\beta$, the reduction is carried out by *conjugating* using a certain ambient isotopy. Next, when reducing to the case $T_\alpha(J)=T'\alpha(J)$, the reduction is again carried out by *conjugating* using a certain ambient isotopy. Finally, when reducing to the case that $T\alpha \mid J = T'_\alpha \mid J$, the reduction is, once again, carried out by *conjugating* using a certain ambient isotopy. – Lee Mosher Feb 24 '25 at 17:18
  • I think I don't understand this reduction fully. If $N_{\alpha}$ and $N_{\beta}$ are ambient isotopic, then there exists an ambient isotopy $h$ such that $h(N_{\alpha},1) = N_{\beta}$. You then went on to show that $T_{\beta}$ is isotopic to $T_{\alpha}^{'}$ which is fine. In this whole discussion I can't see how $N_{\alpha} = N_{\beta}$ is implied or used anywhere. – Dwaipayan Sharma Feb 24 '25 at 18:44
  • I added a few details to clarify the proof that both of $T_\alpha$ and $T'\alpha$ restrict to the identity map on $S - \text{interior}(N\alpha)$. Keep in mind, at that point of the argument we have built an isotopy between $T_\beta$ and $T'\alpha$; from that point onward, the rest of the proof is concerned with building an isotopy between $T'\alpha$ and $T_\alpha$. – Lee Mosher Feb 24 '25 at 19:34
  • Sorry but can you please add some details to the equality of $T_{\alpha}$ and $T_{\alpha}^{'}$ on J? I just cannot understand it. Otherwise the whole proof I have understood. –  Feb 24 '25 at 20:12
  • It is another conjugacy by isotopy argument. Each of these conjugacy by isotopy arguments will look, almost line by line, like the argument that I just updated for proving that $T_\alpha$ and $T'\alpha$ restrict to the identity map in $S - \text{interior}(N\alpha)$. Perhaps you can figure it out? – Lee Mosher Feb 24 '25 at 20:19