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!

P.S. I have asked this at math stack exchange but thought I would ask at this site as well.

  • $\begingroup$ Link: math.stackexchange.com/q/152565/264 $\endgroup$ Jun 2 '12 at 16:38
  • $\begingroup$ "collection of pairwise disjoint isotopy classes of simple closed curves" should probably be "collection of isotopy classes of simple closed curves with pairwise disjoint representatives". Perhaps you also intend the classes to be distinct. This last assumption will be important when you allow arbitrary sign. $\endgroup$
    – Sam Nead
    Jun 2 '12 at 16:44
  • $\begingroup$ @Sam: Yip, I mean the isotopy classes are distinct. Thanks for the clarification. $\endgroup$
    – Sean
    Jun 2 '12 at 17:04
  • $\begingroup$ The question as it stands is not well posed. Certainly I can find representatives $\beta,\beta'$ in their isotopy classes that are in minimal position, and I can also find representatives that are not. So in order for the question to make any sense, you need to describe specific representatives $\beta,\beta'$ and then pose a question about those representatives. $\endgroup$
    – Lee Mosher
    Jun 2 '12 at 17:21
  • $\begingroup$ I don't think you really want this question to be tagged 'topological groups'; 'gt.geometric-topology' and 'gr.group-theory' would probably be appropriate. $\endgroup$
    – HJRW
    Jun 2 '12 at 17:57

First (and trivially), yes, there are representatives $b$ and $b'$ of the curve and its image that are in minimal position. But you aren't really asking this. Correct me if I am wrong, but you are actually asking "Is there a small motion of the image of $b$ placing it in minimal position to the original position of $b$?" Said original position is assumed to be minimal with respect to the curves $\{a_i\}$.

The answer is "no". There is no such small motion. To see this, consider such $b$ and $\{a_i\}$ so that there are two components of $b - n(\cup a_i)$ that are parallel as proper arcs in $S - n(\cup a_i)$.

  • $\begingroup$ What does $b-n(\bigcup a_i)$ mean? $\endgroup$
    – Sean
    Jun 2 '12 at 18:03
  • $\begingroup$ I am guessing that $n()$ is a tubular neighborhood, and $b - n()$ is the complement in $b$ of said tubular neighborhood. $\endgroup$
    – Igor Rivin
    Jun 3 '12 at 2:33
  • $\begingroup$ Yes - sorry I didn't make that clear in the post. $\endgroup$
    – Sam Nead
    Jun 3 '12 at 11:27

Ivanov gives a formula for the case where the exponents are arbitrary integers. It is Lemma 4.2 on page 39 of his book Subgroups of Teichmuller Modular Groups. He says "natural numbers" in the statement, but he means "integers".

  • $\begingroup$ I should add that I am reading the English translation of Ivanov's book, and the typo is in this translation. $\endgroup$ Jul 11 '12 at 20:19

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