from Dehn twists to surgery diagram [closed]

Question

Assume the relation $(ab)^6=1$, for $a$ and $b$ Dehn twists about the meridian and the longitude of a torus. Now if we glue the two ends of $T\times I$ together by either the diffeomorphism $(ab)^6$ or the identity map we should get the same three manifold $T^3$. I know there are ways to get from one surgery diagram to another,when we are getting from one Dehn twist presentation of an element of mapping class group to another presentation, by an algebraic process.(for instance: http://www.ams.org/journals/tran/1992-331-01/S0002-9947-1992-1065603-2/S0002-9947-1992-1065603-2.pdf).

My guess is for relations like the above there is no algebraic process (we can check that the two maps are isotopic but not by an algebraic process), so I was wondering if there is a "canonical way" that the isotopy between the two words as above (as elements of mapping class group not $\pi_1(T) $) tells us how to change between their corresponding surgery diagrams.

Answer 1

EDIT: My previous answer had the incorrect orientation for the Dehn twist about $\beta$. This was courteously pointed out by Mark Bell's comment below.

Let's set some notation to be more precise. Assume $M$ is a torus bundle $T^2\times I/ f$, such that $(x,0)\sim (f(x),1)$. Say $\alpha,\beta$ are the meridian and longitude of the torus respectively, and that $a,b$ generate $\pi_1(T^2)$. Now, let's say that $d_\alpha, d_\beta$ represent Dehn twists about $\alpha$ and $\beta$ respectively. The question is what do $d_\alpha, d_\beta$ do to the generating set $a,b$?

This can be easily computed $d_\alpha(a)=a$ and $d_\alpha(b)=ab$ since a Dehn twist of $\alpha$ about $\alpha$ has now effect and $\alpha$ is a representative of the homotopy class, $a$. However, $\beta$ Dehn twisted about $\alpha$ would result in $\alpha \beta$ or $ab$. Likewise, $d_\beta(a)=ab^{-1}$ and $d_\beta(b)=b$.

We can convert $d_\alpha, d_\beta$ to representatives in $SL(2,\mathbb{Z})$, by:

$d_\alpha \mapsto \pmatrix{1 &1\\0 &1}$ and $d_\beta \mapsto \pmatrix{1 &0\\-1 &1}$.

$w=(d_\alpha d_\beta)^6 \mapsto \pmatrix{0 & 1 \\ -1 &1}^6 = \pmatrix{1 & 0\\ 0 & 1}$.

(I think this is the type of algebraic process that is being asked about.)

The torus bundle should be determined up to orientation preserving homeomorphism by the conjugacy class the matrix in $SL(2,\mathbb{Z})$ that represents $w$. If we allow orientation reversing homeomorphism that the torus bundle is determined by the $SL(2,\mathbb{Z})$ conjugacy classes of $w$ and $w^{-1}$. We can compare the actions of this matrix to the mapping tori carefully described in Scott's Geometries of 3-manifolds in order to determine the manifold up to isotopy. Another good reference is the Twister program, which handles many of the computations for you.