How to rigorously prove that simple closed curves on a surface are primitive closed curves ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:11:25Z http://mathoverflow.net/feeds/question/79929 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79929/how-to-rigorously-prove-that-simple-closed-curves-on-a-surface-are-primitive-clos How to rigorously prove that simple closed curves on a surface are primitive closed curves ? Analysis Now 2011-11-03T14:42:56Z 2011-11-03T22:24:44Z <p>Let me first state the definitions :</p> <p>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 $n\ge 2$.</p> <p>My question is : rigorously prove that simple closed curves are primitive . </p> <p>It is visually pretty clear , but I have difficulty proving it. Thanks !</p> http://mathoverflow.net/questions/79929/how-to-rigorously-prove-that-simple-closed-curves-on-a-surface-are-primitive-clos/79985#79985 Answer by Sam Nead for How to rigorously prove that simple closed curves on a surface are primitive closed curves ? Sam Nead 2011-11-03T22:24:44Z 2011-11-03T22:24:44Z <p>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 $\langle \gamma \rangle &lt; \pi_1(X)$. 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. </p> <p>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. </p>