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.


closed as unclear what you're asking by Ryan Budney, Daniel Moskovich, Stefan Kohl, Neil Strickland, Ricardo Andrade Feb 27 '14 at 21:27

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    $\begingroup$ I don't understand the question. What do you mean by an 'algebraic process'? And I don't want to find out by downloading an entire PDF file. Please just link to the abstract. $\endgroup$ – HJRW Feb 27 '14 at 5:54
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    $\begingroup$ The linked paper is Ning Lu's "Elementary proof of the fundamental theorem of Kirby calculus". I'm not sure what an "algebraic process" means, but you can always translate a surgery description to a Heegaard splitting description, and vice versa. Please clarify what you are asking. $\endgroup$ – Daniel Moskovich Feb 27 '14 at 6:38
  • $\begingroup$ what I meant by algebraic process is verifying algebraically that two words present the same element. $\endgroup$ – nikita Mar 3 '14 at 14:32

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.

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    $\begingroup$ Neil, because of the orientation of $b$, the left Dehn twist $d_\beta$ sends $a$ to $ab^{-1}$ not $ab$. Therefore its matrix in $SL(2, \mathbb{Z})$ is $(1, 0, -1, 1)$. Hence $w$ corresponds to $(0, 1, -1, 1)^6 = (1, 0, 0, 1)$. So $M$ should have Euclidean geometry. By using Alexander's trick on $a \cup b$ you can even check that $(d_\alpha d_\beta)^6 = id$ by hand. Or if you prefer, software packages such as Twister can build the bundle for you and you can check that it is $T^3$. $\endgroup$ – Mark Bell Feb 27 '14 at 8:23
  • $\begingroup$ Mark, thanks for pointing out that error and the helpful comment. I made an edit to correct my previous mistake. $\endgroup$ – Neil Hoffman Feb 27 '14 at 11:27

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