We want to prove that:
Dehn twist is not $\displaystyle{1_{G}}$ in $\displaystyle{MAP(S)}$ and is infinite order.
So, we want to show that: Dehn twists are non-trivial infinite order elements in $\displaystyle{MAP(S)}$.
I want help to proof the above.
First of all we give the following definitions:
Let $S$ be a connected, orientable surface. A continuous map $\displaystyle{\gamma : \mathbb{S}^{1} \longrightarrow S}$ is called a closed curve. It is called simple when we do not have any self intersections in the loop besides the base point.
Given an oriented closed curve $\displaystyle{\gamma \subseteq S}$, we have a bijective correspondence: \begin{align*} \left \{ \begin{matrix} \text{Non-trivial conjugacy} \\ \text{classes in } \pi_{1} ( S ) \end{matrix} \right \} \longleftrightarrow \left \{ \begin{matrix} \text{Non-trivial} \\ \text{free homotopy classes of} \\ \text{oriented closed curves in S} \end{matrix} \right \} \end{align*}
By a free homotopy class we mean homotopy of loops without considering the base point.
Since our surface is connected, we will be able to conjugate the free homotopy loops to the fixed homotopy loops by the path joining the base point to a point in the free homotopy loop. In other words, a loop based at $\displaystyle{x_{1}}$ can be seen as a loop based at $\displaystyle{x_{0}}$ by conjugating it with the path connecting $\displaystyle{x_{0}}$ and $\displaystyle{x_{1}}$. Thus, if homotopic we get the correspondence. Moreover, we have another bijective correspondence: \begin{align*} \left \{ \begin{matrix} \text{elements of the conjugacy} \\ \text{class of } \gamma \in \pi_{1} ( S ) \end{matrix} \right \} \longleftrightarrow \left \{ \begin{matrix} \text{lifts of these curves} \\ \text{free homotopy classes of } \\ \text{to the universal cover of } S \end{matrix} \right \} \end{align*}
Let $\alpha$ and $\beta$ be two closed curves in $S$. Let the free homotopy class of $\alpha$ be $a$ and that of $\beta$ be $b$. We define the minimum number of intersection points between the representatives of the classes $a$ and $b$, i.e. $\displaystyle{i ( a , b ) = \min \big \{ | \alpha \cap \beta | : \alpha \in a , \beta \in b \big \} \geqslant 0 }$ to be the geometric intersection number of $a$ and $b$. Observed that $\displaystyle{i ( a , a ) = 0}$. The representative curves $\displaystyle{\alpha \in a}$ and $\displaystyle{\beta \in b}$ are said to be in minimal position if $\displaystyle{i ( a , b ) = i ( \alpha , \beta )}$.
We refer to a simple closed curve as \textbf{separating} if cutting the surface along the curve breaks the surface into connected components and if cutting the surface along the curve does not result in breaking the surface into connected components we call it nonseparating.
Let $S$ be a surface and $\displaystyle{Homeo(S)}$ be the group of all homeomorphisms of $\displaystyle{S \longrightarrow S}$. Then $Homeo^{+}(S, \partial S)$ is the subgroup of $Homeo(S)$ containing all orientation-preserving homeomorphisms of $S$ to $S$ which are identity on the boundary and preserve the set of punctures.
Let $S$ be a connected, orientable surface. The mapping class group of $S, MAP(S)$ is defined to be the group of the path components of $Homeo^{+}(S, \partial S)$, i.e. $\displaystyle{MAP(S) = \pi_{0} \big( Homeo^{+} ( S , \partial S ) \big)}$. Let $\displaystyle{Homeo_{0} ( S , \partial S )}$ be the path component of identity in $\displaystyle{Homeo^{+} ( S , \partial S )}$ then equivalently, $\displaystyle{MAP(S) = Homeo^{+} ( S , \partial S ) / Homeo_{0} ( S , \partial S ) = Homeo^{+} ( S , \partial S ) / \sim}$ where $"\sim"$ is the isotopy relation.
Let $S$ be a connected oriented surface. Every simple closed curve in $S$ has a regular annular neighbourhood in $S$. Let $\alpha$ be a simple closed curve in $S$ and let $N$ be the annular neighbourhood. Let a homeomorphism $\displaystyle{\varphi : A \longrightarrow N}$ and consider $\displaystyle{T : A \longrightarrow A}$ by the follwig $\displaystyle{T ( \vartheta , t ) = \big( \vartheta + 2 \pi t \text{ , } t \big)}$. The Dehn twist about $\alpha$ is given by $\displaystyle{T_{\alpha} : S \longrightarrow S}$ such that $\displaystyle{T_{\alpha} ( x ) = \begin{cases} \varphi \circ T \circ \varphi^{-1} ( x ) , x \in N \\ \\ x , x \in S \setminus N \end{cases}}$.
We are proving the following statement:
Thereom: Dehn twists are non-trivial elements in the mapping class group of a surface.
Below is my approach: Let $b$ be a non-separating simple closed curve on the surface. By the change of coordinates principle we can bring $b$ to a homology generator say $b_{1}$.
If we find a curve $\alpha_{1}$ such that $\displaystyle{i ( \alpha_{1} , b_{1} ) = 1 \Longrightarrow i ( \alpha_{1} , b) = 1 \Longrightarrow i \big( T_{b} ( \alpha ) , \alpha \big) = 1}$ where $\displaystyle{i ( \alpha , \alpha ) = 0}$. Thus, it is a non-trivial element(we are done).(The argument for the separating curve is similar).
My question is: How can we find this curve $\alpha_{1}$? Can you proof to me why can we find this $\alpha_{1}$? Thank you in advance.
