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Let $M$ be an orientable surface$($without boundary$)$, possibly non-compact. Let $\alpha\in \pi_1(M)$ be a primitive element i.e. $\alpha$ is not a proper power of some other element. Let $f:\Bbb S^1\to M$ be any loop representing $\alpha$. Consider the lifting problem enter image description here

Here, $M_\alpha$ be the cover corresponding to the subgroup $\langle\alpha\rangle$ of $\pi_1(M)$ and $\widetilde M$ be the universal cover of $M$. Also, all unleveled maps are covering maps.

Note that $\pi_1(M)$ acts on $\widetilde M$ via covering transformations, and $M_\alpha$ be an open annulus$($any open connected surface with a finitely generated fundamental group is homeomorphic to the interior of a closed surface$)$.

$\textbf{Problem 1:}$ Consider two sets $$\mathscr A_f:=\big\{g\in \pi_1(M):\text{ the map } \Bbb R\xrightarrow{g\cdot \ell}\widetilde M\to M_\alpha\text{ runs from one end to other of the open annulus }M_\alpha\big\},$$ $$\mathscr B_f:=\big\{(z,w)\in \Bbb S^1\times\Bbb S^1:z\not= w\text{ and }f(z)=f(w)\big\}$$ Is the cardinality of $\mathscr B$ two-times the cardinality of $\mathscr A$, i.e. $|\mathscr A_f|=\frac{1}{2}|\mathscr B_f|$?

$\textbf{Problem 2:}$ Give a Riemannian metric on $M$ such that $\alpha$ can be represented by a shortest loop $f^\#:\Bbb S^1\to M$, i.e. $f^\#$ has the minimum length in its free homotopy class. Is $|\mathscr A_f|\geq |\mathscr A_{f^\#}|$?

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For problem one, the answer is "no" as there may be "bigons". The pair of intersections forming the bigon contribute to $B_f$ but not to $A_f$.

For problem two, the answer is "yes" as the shortest representative has no bigons.

A version of this is discussed in the "Primer on mapping class groups" by Farb and Margalit under the name "the bigon criterion".

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  • $\begingroup$ Another question does every(compact or non-compact) surface without boundary has the property: there is a Riemannian metric on the surface such that every non-trivial element of the fundamental group can be represented by a shortest loop. $\endgroup$
    – Someone
    Jan 30, 2021 at 9:39
  • $\begingroup$ How do I relate the words "shortest", "bigons', "$\mathscr A_f$? $\endgroup$
    – Someone
    Jan 30, 2021 at 9:44
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    $\begingroup$ For your first question in the comments - there is always a constant curvature metric that does the job. For your second question in the comments - I recommend you look at the reference I suggested. $\endgroup$
    – Sam Nead
    Jan 30, 2021 at 9:50
  • $\begingroup$ Sorry, but any reference for "there is always a constant curvature metric that does the job" I am asking because most of the Riemannian geometry book deals with the compact case. Also, I have read the proof bigon-criterion(second proof), and there is no link(as far as I can see) it with the shortest. $\endgroup$
    – Someone
    Jan 30, 2021 at 9:58
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    $\begingroup$ The compact and non-compact cases are dealt with in the same way - this is an application of the uniformisation theorem. As for your second question - prove that a shortest representative cannot have bigons. [Hint - draw a picture.] $\endgroup$
    – Sam Nead
    Jan 30, 2021 at 14:44

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