Isotopy between overlapping and non-over. Dehn Twists.

[Bacle]

Hi, everyone:

I am curious about this: two Dehn twists D1,D2, commute if the annuli A1, A2
are disjoint, so that the action of D1 does not affect D2 ( and the respective curves
C1,C2 about which we do the twists.)

Now, there is also a result that if the annuli A1, A2 , _do_ overlap, that
D2oD1 is not, of course, equal to D2oD1 if/when A1,A2 are disjoint, but, we
do have that the two twists are isotopic, i.e., the twist D2oD1 , with A1/\A2={}
is isotopic to the twist D1oD2 , when A1/\A2 =/ {}.

Anyone have an idea for how to show this.? . Sorry I don't have Rolfsen
(nor his book : ) ) available at the moment.

Thanks in Advance.

 

[quasar987]

Um, I will try, but this kind of thing is so much easier to explain on a blackboard. For simplicity, assume D1 and D2 both take place along the "small diameter of the torus" (much easier to visualize) and also assume that both D1 and D2 are 1-turn twists in the same direction!

First, set A3=A1 n A2. Then consider the homotopy that first "stretches" A3 to a big annuli A3' covering like half the torus. Then as a curve goes around the big diameter of the torus, it slowly twists twice around the small diameter as it passer through the region A3'. Select a sub annuli A4 of A3' and homotope the region A3' so that at time t=1 of the homotopy, our traveling curve does its first twists.., then is constant on A4... then does its second twist. Set one of the connected component of A3' \A4 to be A1' and the other to be A2'. Then the resulting self homeo of T^2 at time t=1 of the homotopy is a dehn twist in A1' following by a Dehn twist on A2' as required...