Mapping class group and cylindrical structure

2012-02-16T08:08:03Z

Let us fix a torus $\Sigma=S^1 \times S^1$. We consider a cylinder $\Sigma \times I$ and a data $(\Sigma\times I, \Sigma\times 0, \Sigma\times 1, f_{0},f_{1})$. Here $f_{i}$, called parametrization, is a homeomorphism from $\Sigma$ to $\Sigma \times i$, for $i=0, 1$. Two such data are equivalent if there is a homeomorphism $\Sigma \times I$ to itself such that on the boundary it commutes with parametrizations.

Let us consider the composition

$H_1(\Sigma; \mathbb{Z}) \to H_1(\Sigma \times 0; \mathbb{Z}) \to H_1(\Sigma \times 1; \mathbb{Z}) \to H_1(\Sigma; \mathbb{Z})$.

Here the first and the third ismorphism are induced by the parametrizations $f_{i}$, $i=0, 1$ respectively and the second isomorphism (let's say $h$) is obtained by pushing loops in the bottom base of $\Sigma \times I$ to the top base using the cylindrical structure on $\Sigma \times I$.

Note that $h$ might not be the identity and might differ by each data.

The composition $f_{1}^{-1}hf_{0}$ of these gives an element of a mapping class group $M(\Sigma)$ of $\Sigma$.

Question: If two data induce the same element in $M(\Sigma)$, are these two data equivalent? Can you give a concrete homeomorphism $\Sigma \times I$ to itself that gives an equivalence?

Answer by Misha for Mapping class group and cylindrical structure

2012-03-03T23:35:17Z

It appears that you are asking if concordance of homeomorphisms of tori is equivalent to their isotopy. (Your notion of "equivalent data" simply means "concordant homeomorphisms", your definition of the mapping class group for the tori is equivalent to the standard one since your surface has genus 1.) The answer is yes. If $\Sigma$ is the torus then two homeomorphisms of $\Sigma$ are isotopic iff they induce the same map of $H_1(\Sigma)$. (False in the higher genus case.) What you observed for tori (using homology), is that concordance of $f_1, f_2$ implies that they induce the same map of $H_1(\Sigma)$ which, in turn, implies that $f_1, f_2$ are isotopic. The converse (which, I think, is your question) is obvious: If $f_t$ is an isotopy of $f_1$ to $f_2$ then the map $F: \Sigma\times I\to \Sigma\times I$ given by

$$F(x,t)=f_t(x)$$

is the required concordance. Actually, the same conclusion Concordance $\iff$ isotopy holds for higher genus surfaces as well. The nontrivial direction is Concordance $\Rightarrow$ isotopy. The reason it holds is a theorem by Waldhausen (although, in the special case when $M=\Sigma \times I$, the result is due to Stallings, I think):

Theorem. Suppose that $M$ is a Haken manifold, $f_1, f_2: M\to M$ are homeomorphisms which agree on the boundary. Then $f_1$ is isotopic to $f_2$ (rel. boundary) iff $f_1, f_2$ induce the same map of the fundamental group.

Mapping class group and cylindrical structure - MathOverflow

Mapping class group and cylindrical structure

Answer by Misha for Mapping class group and cylindrical structure