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I am interested in the following problem : decide if a certain element of the fundamental group can be represented by a simple closed curve. The general case has already been asked and answered on MO (see How to detect a simple closed curve from the element in the fundamental group?).

I am interested in a more specific case :

To make things simpler, let us assume we are dealing with a genus $2$ oriented closed surface with its usual set of generator of its fundamental group $a_1$, $b_1$, $ a_2$, $b_2$. It is clear that the commutator $[a_1,b_1]$ can be represented as a simple closed curve : take the curve separating $S$ in two $1$ holed torus in the classic presentation.

My question is the following : is there any other commutator on the free subgroup generated by $a_1$ and $b_1$ that can be representated by a simple closed curve ? My attempts to find such a curve all failed, checking on simple ones like $[a_1, b_1^2]$, so I wondered whether it was possible or not.

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The free subgroup generated by $a_1,b_1$ is carried by a one-holed torus $T$ embedded in the surface. Let $\gamma$ be a simple closed curve embedded in $T$, and consider $T\smallsetminus \gamma$. A simple argument with Euler characteristic shows that:

  • either $T\smallsetminus\gamma$ is the union of a one-hold torus and an annulus, so $\gamma$ was boundary parallel and hence conjugate to $[a_1,b_1]$; or

  • $T\smallsetminus\gamma$ is a thrice-punctured sphere, so $\gamma$ was non-separating.

In conclusion, every simple closed curve in $\langle a_1,b_1\rangle$ is either conjugate to $[a_1,b_1]$ or non-separating.

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  • $\begingroup$ Just to clarify, the term "conjugate" here means not that they are conjugate as elements of the fundamental group, but that they are equivalent under some automorphism of the fundamental group. The same argument shows for general $g$ that every simple closed curve in the commutator subgroup is equivalent to $[a_1,b_1]\cdots[a_k,b_k]$ for some $1\leq k<g$. $\endgroup$
    – Tom Church
    Dec 3, 2013 at 17:29
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    $\begingroup$ @TomChurch : In this case, conjugate really means conjugate. The point is that there is only one homotopy class of non-nullhomotopic simple closed separating curve on a one-holed torus. When you fix a base point, this picks out a single conjugacy class. In higher genus, of course, you need general automorphisms. $\endgroup$ Dec 3, 2013 at 19:46
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    $\begingroup$ (another observation to make this more clear : the mapping class group fixes the homotopy class of a boundary parallel curve on a surface with boundary) $\endgroup$ Dec 4, 2013 at 1:50
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    $\begingroup$ You're completely right -- thanks, Andy! $\endgroup$
    – Tom Church
    Dec 7, 2013 at 15:24

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