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I have a question about a proof that I am reading in "A primer on Mapping Class Groups" by Farb and Margalit.

Let $a$ be a simple closed curve in a compact surface $S$ (possibly with marked points and boundary components) not isotopic to a point or boundary component and let $T_a$ denote the Dehn twist about $a$.

Now let $\alpha_1 , ..., \alpha_n$ be a collection of pairwise disjoint isotopy classes of simple closed curves in $S$ and let $M = \prod_{i=1}^{n} T_{\alpha_i}^{e_i}$. Also suppose that $e_i >0 $ $\forall i$ or $e_i <0$ $\forall i$ where $e_i$ is an integer and that $b$ is an arbitrary isotopy class of a simply closed curve.

Ok now we look at $M(b)$, and find a representative $\beta '$ in its isotopy class. We also take $\beta$ to be in the isotopy class of $b$.

  • I now want to show that $\beta$ and $\beta'$ are in minimal position!

For context, this is propostion 3.4 in Farb and Margalit's "A primer on Mapping Class Groups".

If one lets $n=1$ above then it wouldn't be hard to prove (prop 3.2 in the same book) as it follows from the bigon criterion. However they say that it also is true here for $\beta$ and $\beta'$ since all the $e_i$'s have the same sign or rather all the twists are in the same direction.

  • So what happens if we allow the $e_i$'s to have arbitrary sign?

Any comments/help would be greatly appreciated! cheers!

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Posted on MO: mathoverflow.net/questions/98666/… –  Zev Chonoles Jun 2 '12 at 16:38
    
@ Zev. Thanks for the link. –  Sean Jun 2 '12 at 16:58

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