Let $M$ be an orientable surface equipped with a Riemannian metric, and $\alpha\in \pi_1(M)$ be a primitive element. Let $f$ be a shortest loop representating $\alpha$ and $p_\alpha:M_\alpha\to M$ be the covering corresponding to the subgroup $\langle \alpha\rangle$ of $\pi_1(M)$. Let $p:\widetilde M\to M$ be the universal cover.
Consider the lift $f_\alpha:\Bbb S^1\to M_\alpha$ be the lift of $f$ w.r.t. the covering $p_\alpha:M_\alpha\to M$. Now, consider the lift $\ell:\Bbb R\to \widetilde M$ of the composite map $\Bbb R\xrightarrow{\exp}\Bbb S^1\xrightarrow{f_\alpha}M_\alpha$.
Problem 1: Show that $\ell:\Bbb R\to \widetilde M$ is a proper map.
Problem 2: Consider the action $\bullet$ of $\pi_1(M)$ on $\widetilde M$ via deck-transformations. For any $g\in \pi_1(M)$, the composition $\Bbb R\xrightarrow{g\bullet \ell}\widetilde M\xrightarrow{\text{universal cover}}M_\alpha$ is a proper map.
$(1)$ I am trying to use use the fact $[f_\alpha\circ \gamma]\bullet \ell(x)=\ell\big([\gamma]\bullet x\big)$ with the fact that the action of the set of all deck transformations(for the covering $p:\widetilde M\to M$) on $\widetilde M$ is properly discontinuous.
$(2)$ For this I need to use fiber of covering is discrete and $g\bullet\ell$ is an embedding.
I can not complete the proof. Any help will be appreciated. Thanks, in advance.
Below is a screenshot of the book An Introduction to 3-manifolds by Peter Scott
Best Answer
Suppose that $X$ is a complete (connected) Riemannian manifold, $\Gamma$ is a group of isometries of $X$ acting on $X$ properly discontinuously.
Definition. Let $\gamma\in\Gamma$ be an element of infinite order. Then a complete geodesic $A_\gamma\subset X$ is called an {\em axis} of $\gamma$ if it is invariant under $\gamma$.
(In this case, $\gamma$ acts on $A_\gamma$ as a translation. Note that an axis need not be unique.)
Given an infinite order element $\alpha\in\Gamma$, let $X_\alpha$ denote the quotient Riemannian manifold $X/\langle \alpha\rangle$.
I will be always parameterizing geodesics by the arc-length.
Lemma 1. Suppose that $\alpha, \beta\in\Gamma$ are elements of infinite order so that the subgroups $\langle \alpha\rangle$, $\langle \beta\rangle$ have trivial intersection. Then the projection $A_\beta\to X_\alpha$ is a proper map.
Suppose not. Then there exists a sequence of points $z_i\in A_\beta$ diverging to infinity such that the projections of these points to $X_\alpha$ are within distance $R$ from some point $\bar{x}\in X_\alpha$ (projection of $x\in X$).
Since $\langle \beta\rangle$ acts cocompactly on the axis $A_\beta$, there is a constant $C$ and a diverging sequence of integers $n_i$ such that the (minimal) distance from $\beta^{n_i}(x)$ to the $\langle\alpha\rangle$-orbit of $x$ is $\le C$. In other words, there is a diverging sequence of integers $(m_i)$ such that $$ d(\beta^{n_i}(x), \alpha^{-m_i}(x))\le C. $$ Without loss of generality, we may assume that both sequences consist of distinct integers. We have $$ d(\alpha^{n_i} \beta^{n_i}(x), x)\le C $$
By the proper discontinuity of the action of $\Gamma$ on $X$, the set of products $\alpha^{n_i} \beta^{n_i}$ is finite. Hence, there exist distinct $i, j$ such that $$ \alpha^{n_i} \beta^{n_i}= \alpha^{n_j} \beta^{n_j}, $$ $$ \alpha^{n_i-n_j}= \beta^{n_j-n_i}, $$ which is a contradiction. qed
Suppose now that $\Gamma$ satisfies the following maximality condition: If $\alpha, \beta$ are primitive elements of $\Gamma$ then the subgroups generated by $\alpha, \beta$ are either equal or have trivial intersection. For instance, fundamental groups of surfaces satisfy this property. (Let me know if you need help proving this property. For noncompact surfaces this follows from the fact that their fundamental groups are free.)
Corollary 1. Suppose that $\alpha\in \Gamma$ is a primitive element. Then for each $\gamma\in\Gamma$ $\alpha, gamma$ either commute or anticommute: $$ [\alpha,\gamma]=1, \hbox{or} \gamma\alpha\gamma^{-1}=\alpha^{-1}, $$ or the geodesic $\gamma(A_\alpha)$ projects properly to $X_\alpha$.
Proof. Consider $$ \beta= \gamma\alpha\gamma^{-1}. $$ Then $A_\beta=\gamma(A_\alpha)$. By Lemma 1, if the projection of $A_\beta$ to $X_\alpha$ is not proper, then the subgroups generated by $\alpha, \beta$ have nontrivial intersection. By the Maximality Property above, these subgroups have to be equal (note that the primitivity of $\alpha$ implies that of $\beta$ since conjugation is a group automorphism). Thus, $$ \langle \alpha\rangle= \langle\gamma\alpha\gamma^{-1}\rangle. $$ Since the infinite cyclic group $\langle \alpha\rangle$ has only two generators, $\alpha^{\pm 1}$, we get
$$ [\alpha,\gamma]=1, \hbox{or} \gamma\alpha\gamma^{-1}=\alpha^{-1}, $$ qed
Suppose now that $X$ is 2-dimensional and $\Gamma$ acts freely. If $\Gamma$ preserves orientation, i.e. $M=X/\Gamma$ is orientable, the anti-commutation is impossible. Furthermore, unless the group $\Gamma$ is isomorphic to ${\mathbb Z}^2$ and the quotient $X/\Gamma$ is diffeomorphic to the torus, the commutation is only possible if $\gamma$ belongs to the subgroup generated by $\alpha$.
Thus, we obtain:
Corollary 2. Suppose that $X$ is 2-dimensional, $M=X/\Gamma$ is orientable and is not diffeomorphic to the torus. Then either $\gamma(A_\alpha)=A_\alpha$ or $\gamma(A_\alpha)$ projects properly to $X_\alpha$.
Note that the torus is a genuine exception: No matter what $\gamma\in \Gamma$ you take, $\gamma(A_\alpha)$ does not project properly to $X_\alpha$ since $\gamma(A_\alpha)$ is invariant under $\alpha$ and, hence, projects to a compact in $M_\alpha$.