algebraic-topologymanifoldsmapping-class-group
Q1. In the book "A primer on mapping class groups", the author gives a definition of essential closed curve as "A closed curve is called essential if it is not homotopic to a point, puncture, or a boundary component."
Being homotopic is a property of two (continuous) maps, so I'm not able to understand this definition.
Also can you provide me an example of an essential closed curve?
Q2. What is the definition of "orientation – preserving" homeomorphism?
Thanks in advance.
This is true not just for $S^2$, but for any manifold of dimension at least $2$. That is,
For any $n\geq 2$, and any $n$-manifold $M$, any permuation of a finite subset $\{x_1, x_2,..., x_k\}\subseteq M$ extends to an orientation preserving diffeomorphism of $M$.
In fact, one can choose this diffeomorphism to be isotopic to the identity - that is what the below proof actually gives. Further, we really do need $n\geq 2$: if $a<b<c\in \mathbb{R}$, there is a no homeomorphism of $\mathbb{R}$ which maps $a$ to $b$, $b$ to $c$ and $c$ to $a$.
I'll prove the above claim via a series of propositions.
Proposition 1: Suppose $M^n$ is a manifold with $n\geq 2$. Choose distinct points $x,y, z_1,..., z_k\in M$. Then there is an orientation preserving diffeomorphism $f:M\rightarrow M$ for which $f(x) = y$, but $f(z_i) = z_i$ for all $1\leq i\leq k$.
Proof: Because $n\geq 2$, $M\setminus\{z_1,..., z_k\}$ is path connected. Let $\gamma:[0,1]\rightarrow M$ be a simple regular curve with $\gamma(0) = x$ and $\gamma(1) = y$. Let $U$ denote a tubular neighborhood of $\gamma$, chosen small enough so that $U\cap\{z_1,..., z_k\} = \emptyset$. Now, create a vector field $X$ extending $\gamma'$ which is supported in $U$. Flowing for an appropriate amount of time gives the desired $f$. $\square$
Now, let $X\subseteq M$ be any finite set. Given a permutation $\sigma$ of $X$, we will say $f$ is an extension of $\sigma$ if $f:M\rightarrow M$ is an orientation preserving diffeomorphism and $f|_X = \sigma|_X$.
Proposition 2: If $\sigma:X\rightarrow X$ is a transposition, there is an extension $g$ of $\sigma$.
Proof: Suppose $\sigma(x_i) = x_j$ and $\sigma(x_j) = x_i$ for $x_i\neq x_j$, with all other $x_k\in X$ being fixed. Applying the lemma once with $x = x_i, y = x_j$ and with $\{z_1,..., z_k\} = X\setminus \{x_i, x_j\}$, we get a diffeomorphism $f_1$ for which $f_1(x_i) = x_j$, but $f_1(x_k) = x_k$ for all other $k$. However, there is no reason that $f_1(x_j) = x_i$. If $f_1(x_j) \neq x_i$, then we use the lemma again with $x = f_1(x_j)$, $y = x_i$ and $\{z_1,..., z_{k+1}\} = X\setminus\{x_i\}$ to get a diffeomorphism $f_2$ with $f_2 f_1(x_j) = x_i$ and $f_2(x_k) = x_k$ for any other $k$. Then $g = f_2\circ f_1$ is the desired diffeomorphism of $M$. $\square$
Proposition 3: Suppose $\sigma_1, \sigma_2$ are both permutations of $X$ with extensions $f_1$ and $f_2$. Then $\sigma_1 \sigma_2$ is extended by $f_1\circ f_2$. Further, $\sigma_1^{-1}$ is extended by $f_1^{-1}$.
Proof: Let $x_i\in X$. For the first statement, $f_2(x_i) = \sigma_2(x_i)\in X$ and so $f_1 (f_2(x_i)) = \sigma_1(f_2(x_i)) = \sigma_1(\sigma_2(x_i))$.
For the second statement, note that $x_i = \sigma_1(\sigma_1^{-1}(x_i)) = f_1(\sigma_1^{-1}(x_i)$, and also that $x_i = f_1(f_1^{-1}(x_i))$. Since $f_1$ is injecive, $\sigma_1^{-1}(x_i) = f_1^{-1}(x_i)$.$\square$
Proposition 3 essentially claims the following: The set $\{\sigma \in S_k: \exists \text{ extension }f\}$ is a subgroup of the symmetric group on $k$ letters, $S_k$. Proposition 2 then claims that this set contains all transpositions. Since the symmetric group is generated by transpositions, the claim above now follows.
I forgot all about the fact that I asked this question here, but I answered it myself a while back.
Definitions:
A surface $S$ is an oriented connected sum of $g \geq 0$ tori with $b \geq 0$ open disks removed, and $n \geq 0$ punctures in its interior. Homeomorphism classes of surfaces are in bijective correspondence with $\{(g,b,n): g,b,n \geq 0\}$.
Given a surface $S$, the group (under composition) of orientation preserving homeomorphisms that fix the boundary $\partial S$ is denoted by Aut$^+(S,\partial S)$. This is endowed with the compact-open topology. The path component of Aut$^+(S,\partial S)$ containing id$_S$ is denoted Aut$_0(S,\partial S)$.
Theorem:
Given a surface $S$, the following are three equivalent definitions of its mapping class group, Mod$(S)$:
- Mod$(S)_1 := \pi_0 (\mathrm{Aut}^+(S,\partial S), \mathrm{id}_S)$.
- $\mathrm{Mod} (S)_2$ is the group of boundary-fixing isotopy classes of maps in $\mathrm{Aut}^+(S,\partial S)$.
- $\mathrm{Mod}(S)_3 := \mathrm{Aut}^+(S,\partial S)/ \mathrm{Aut}_0(S,\partial S)$.
We first show that $1$ and $2$ are naturally isomorphic as sets. Explicitly, $\pi_0 (\mathrm{Aut}^+(S,\partial S), \mathrm{id}_S)$ is the collection of homotopy classes of continuous maps $\sigma: \boldsymbol 2 \to \mathrm{Aut}^+(S,\partial S)$ such that $\sigma (0) = \mathrm{id}_S$. Note that $\boldsymbol 2$ denotes the discrete space $\{0,1\}$. Suppose $\sigma, \tau : \boldsymbol 2 \to \mathrm{Aut}^+(S,\partial S)$ belong to the same homotopy class. Then there is a continuous map $$H: \boldsymbol 2 \times [0,1] \to \mathrm{Aut}^+(S,\partial S)$$ such that $H(1,0) = \sigma(1)$, $H(1,1) = \tau(1)$, and $H(0,t) = \mathrm{id}_S$ for each $t \in [0,1]$. Consider the map $$F: S \times [0,1] \to S$$ defined by $F(s,t) := H(1,t)(s)$ for all $s \in S, t \in [0,1]$. Then $F(s,0) = \sigma(1)(s)$ and $F(s,1) = \tau(1)(s)$ for each $s \in S$. Given any $t \in [0,1]$, $F(-,t) \in \mathrm{Aut}^+(S,\partial S)$. Therefore $F$ is a boundary-fixing isotopy from $\sigma(1)$ to $\tau(1)$. In summary, given any homotopy class $[\sigma] \in \pi_0 (\mathrm{Aut}^+(S,\partial S), \mathrm{id}_S)$, evaluation at $1$ gives a boundary-fixing isotopy class of maps in $\mathrm{Aut}^+(S,\partial S)$.
Conversely, suppose $f,g \in \mathrm{Aut}^+(S,\partial S)$ are isotopic. One can similarly construct a homotopy between the maps $\sigma,\tau: \boldsymbol 2 \to \mathrm{Aut}^+(S,\partial S)$ defined by $\sigma(0)=\tau(0)= \mathrm{id}_S$, $\sigma(1) = f$, $\tau(1) = g$.
We now define the group structure on $\mathrm{Mod} (S)_2$. I claim that $$[f][g] = [f\circ g]$$ for each $[f],[g]\in \mathrm{Mod}(S)_2$ defines a group structure on $\mathrm{Mod} (S)_2$. Let $f_0,f_1,g_0,g_1 \in \mathrm{Aut}^+(S,\partial S)$ such that $[f_0] = [f_1]$ and $[g_0] = [g_1]$. Then there are boundary-fixing isotopies $F^f, F^g: S \times [0,1] \to S$ such that $F^f(s,i) = f_i(s)$ and $F^g(s,i) = g_i(s)$ for all $s \in S$. The map $$F: S \times [0,1] \to S$$ defined by $F(s,t) = F^f(F^g(s,t),t)$ for all $s\in S, t\in [0,1]$ is then a boundary-fixing isotopy from $f_0 \circ g_0$ to $f_1 \circ g_1$. Thus $[f][g] = [f\circ g]$ is well defined.
It is now routine to see that this binary operation is a group operation, where $$[\mathrm{id}_S] = e,\ \text{and}\ [f]^{-1} = [f^{-1}].$$ This also induces a group structure on $\mathrm{Mod}(S)_1$: For any $[\sigma],[\tau] \in \pi_0(\mathrm{Aut}^+(S,\partial S), \mathrm{id})$, $$[\sigma][\tau] = [\gamma],\ \text{where}\ \gamma(1) = \sigma(1) \circ \tau(1).$$
Finally, we determine how definitions $1$ and $2$ of $\mathrm{Mod}(S)$ correspond to $\mathrm{Mod}(S)_3$. $\mathrm{Aut}^+(S,\partial S)$ is naturally equipped with a group structure under composition. We first show that $\mathrm{Aut}_0(S,\partial S)$ is a normal subgroup of $\mathrm{Aut}^+(S,\partial S)$, and then that the quotient group $\mathrm{Aut}^+(S,\partial S)/\mathrm{Aut}_0(S,\partial S)$ is isomorphic to $\mathrm{Mod}(S)_2$.
Observe that there is a one-to-one correspondence between isotopies $F: S \times [0,1] \to S$ and paths $\gamma : [0,1] \to \mathrm{Aut}^+(S,\partial S)$. Given an isotopy $F$, simply define the path by $$t \mapsto (s \mapsto F(s,t)).$$ Thus $\mathrm{Aut}_0(S,\partial S)$ is the isotopy class of $\mathrm{id}_S$. From the above discussion about the group operation on $\mathrm{Mod}(S)_2$, it is now clear that $\mathrm{Aut}_0(S,\partial S)$ is a normal subgroup of $\mathrm{Aut}^+(S,\partial S)$. Thus the quotient is truly a group.
I now claim that $\varphi: \mathrm{Mod}(S)_3 \to \mathrm{Mod}(S)_2$ defined by $f + [\mathrm{id}] \mapsto [f]$ is a group isomorphism. Suppose $f,g \in \mathrm{Aut}^+(S,\partial S)$ such that $f+[\mathrm{id}] = g+[\mathrm{id}]$. Then there exists $h \in \mathrm{Aut}_0(S,\partial S)$ such that $f = g \circ h$. Then $[f] = [g\circ h] = [g][h] = [g][\mathrm{id}] = [g]$, so $\varphi$ is well defined. Clearly the map is surjective and injective. Finally, given any $f+[\mathrm{id}], g+[\mathrm{id}]$, $$\varphi(f+[\mathrm{id}])\varphi(g+[\mathrm{id}]) = [f][g] = [f\circ g] = \varphi (f\circ g + [\mathrm{id}]) = \varphi((f + [\mathrm{id}])(g + [\mathrm{id}])).$$ Therefore $\varphi$ is an isomorphism as required. This proves the equivalence of the three definitions for $\mathrm{Mod}(S)$.
Best Answer
A curve is a continuous map $[0,1]\to M$, where $M$ is the manifold on which you're considering it. In the same way one can define a point (a constant curve) and a boundary component if the manifold is of dimension 2 (and therefore the boundary is of dimension 1, meaning that it is a disjoint union of curves). An example of essential closed curve is given in the following picture taken from Basic results on braid groups by Juan González-Meneses.
That depends first of all on how you define orientations. Since I don't want to assume any smooth structure, I'll say it in terms of local homology. An orientation on a manifold $M$ of dimension $n$ is a choice of generator $[M]_x$ of $H_n(M,M-x;\mathbb{Z})$ for each $x\in M$ in such a way that this choice is continuous (See Definition 3.1, and the others if you want another notion). A homeomorphism is orientation preserving if the induced maps on local homology takes orientation to orientations.