Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let me first state the definitions :

A not-nullhomotopic closed curve / loop $c$ on an orientable surface $X,c:[0,1]\to X$ is called simple closed curve is $c|[0,1)$ is injective and [ $c(0)=c(1) ] ; $ A closed curve / loop $c$ is called primitive if in the fundamental group $\pi_1(X,c(1)),$ the homotopy class $[c]$ can NOT be written as $[c]= [\gamma]^n$ for some closed curve $\gamma$ with $\gamma(0)=\gamma(1)=c(0)=c(1) $ and for some $n\ge 2$.

My question is : rigorously prove that simple closed curves are primitive .

It is visually pretty clear , but I have difficulty proving it. Thanks !

share|improve this question
First, you need to assume that c is essential, so that it doesn't represent the trivial element of \pi_1. Second, you need to be careful, since it's not true on non-orientable surfaces (since the boundary of a Mobius strip is twice the core curve). But once you're orientable, the first homology (abelianization) of the fundamental group is free abelian, and c can be chosen to represent a basis element of this free abelian group. Thus it is primitive in homology, and hence in the fundamental group. –  Daniel Groves Nov 3 '11 at 15:13
Thanks , I have edited my question. But when you say " $c$ can be chosen to represent a basis element of the first homology group ", do you assume that c is a non-separating simple closed curve in X and use the result that given any two non-separating simple closed curves $c1,c2$ in an oriented surface $X$ , there exists a global homeomorphism of the surface carrying $c1$ onto $c2$, and then taking $c_1$ as $c$ and $c_2$ and generator of $H_1(X)$ ? Is there a similar result for separating closed simple curves ? –  Analysis Now Nov 3 '11 at 15:26
There's a proof in Farb and Margalit's "Primer on mapping class groups". –  Richard Kent Nov 3 '11 at 15:32
Oh right, I was assuming separating. But as Richard says, there's a proof in the non-separating case in Farb and Margalit's book... –  Daniel Groves Nov 3 '11 at 15:33
add comment

1 Answer

up vote 7 down vote accepted

Suppose that $c = \gamma^n$ in $\pi_1(X)$. Note that, as $\pi_1(X)$ is torsion free and $c$ is assumed to be non-trivial, the element $\gamma$ generates an infinite cyclic subgroup $\langle \gamma \rangle < \pi_1(X)$. Let $A = X^\gamma$ be the cover of $X$ corresponding to the subgroup $\langle \gamma \rangle$. So $\pi_1(A)$ is also infinite cyclic. Since $X$ is orientable, so is $A$. It follows from the classification of surfaces $A$ is a (non-compact) annulus.

Note that $\gamma$ can be lifted to $A$ and this lift, $\gamma'$, is homotopic to the core curve of $A$. Likewise $c$ lifts to a curve $c'$ and we have $c' = (\gamma')^n$ in $\pi_1(A)$. Since $c$ is simple in $X$ the lift $c'$ is simple in $A$. By the intermediate value theorem (sort of!) the only simple curves in $A$ are isotopic to the trivial curve and to the core curve. Thus $n = \pm 1$ and we are done.

share|improve this answer
Sam Nead : thanks for the helpful answer ! I just have a quick question about the last line : I understand that $c$ is isotopic to the core curve of $X_\gamma$, since it is non-trivial. But then how exactly are we saying that $n=+/−1$ ? Are we considering the winding number of $c'$ and $\gamma'$ and getting that 1=n. (winding number of $\gamma$),which means n=1 ? And what is the version of intermediate value theorem that you are using here , that proves : simple closed curves in the annulus are either trivial or isotopic to the core curve ? –  Analysis Now Nov 17 '11 at 17:06
If two curves are isotopic in a surface then (up to conjugacy) they are the same or inverses in the fundamental group. This is a matter of orientation of the curves. When I refer to the "intermediate value theorem" I am giving a three word sketch of the classification of simple closed curves in the annulus. That classification is an exercise in combinatorial topology. –  Sam Nead Nov 17 '11 at 18:00
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.