# Representation of elements of fundamental group by shortest loops

I have some questions from the following text: $$\bullet$$ It is clear to me that $$f_\alpha$$ is a loop and represents a generator of $$\pi_1(M_\alpha)$$ as covering induces injective map on fundamental groups, but why $$f_\alpha$$ is shortest?

Recall that a loop on a surface equipped with a Riemannian metric is called shortest if it has the shortest length in its free homotopy class.

$$\bullet$$ I think here $$l:\Bbb R\to \widetilde M$$ is obtained as the lifting of the following composition map $$\Bbb R\xrightarrow{\exp}\Bbb S^1\xrightarrow{f_\alpha} M$$ w.r.t. the universal covering $$p:\widetilde M\to M$$ using the fact $$\Bbb R$$ is simply-connected. Now, $$l$$ is simple as $$f_\alpha$$ is simple. Am I right?

$$\bullet$$ Why the map $$l:\Bbb R\to \widetilde M$$ is properly embedded in $$\widetilde M$$? I don't know the meaning of "proper" here. I think the term embedding comes to form the fact that $$l$$ is simple.

$$\bullet$$ What will be the full pre-image of $$f_\alpha(\Bbb S^1)$$ in universal cover i.e. what is $$p^{-1}\big(f_\alpha(\Bbb S^1)\big)$$? I guess it will be $$\pi_1(M)$$-tanslates of $$l$$, but have no proof.

Any help will be appreciated. It will be better if someone tells me some text containing all these basic facts.

• Seems we might need to know lemma 5.2.1. Can you give a reference for this (book and location)? Jan 17, 2021 at 19:00
• Lemma 5.2.1 Let $A$ be an open annulus with a Riemannian metric. Let $\gamma$ be a shortest loop on $A$ representing a generator of $π_1(A).$ Then, $\gamma$ is simple. Also, if $\gamma'$ a second such loop on $A$, then $\gamma'$ equals $\gamma$ or is disjoint from $\gamma$. This is on page 101. Jan 17, 2021 at 19:14

In the first bullet, suppose that $$M_\alpha$$ contained a loop $$f'_\alpha : \mathcal S^1 \to M_\alpha$$ in the same free homotopy class as the lift $$f_\alpha$$ but with length shorter than $$f_\alpha$$. Then the loop $$f' : \mathcal S^1 \to M$$ obtained by composing $$f'_\alpha$$ with the covering map $$M_\alpha \mapsto M$$ would be a shorter loop in the same homotopy class as $$f$$, which is a contradiction (the homotopy from $$f_\alpha$$ to $$f'_\alpha$$, when composed with the covering map $$M_\alpha \mapsto M$$, would be a homotopy from $$f$$ to $$f'$$).

In the second bullet, some care is needed. First one needs to verify that $$\alpha$$ has infinite order, but this is true in the fundamental group of any oriented surface: every nontrivial element has infintie order. Also, however, your interpretation does not quite make sense because the statement says that $$l$$ is the preimage of $$f_\alpha(S^1)$$, so $$l$$ itself is a set, not a function. However, if instead you state your interpretation to say that $$l$$ is the image in $$\widetilde M$$ of the map $$\mathbb R \mapsto \widetilde M$$ obtained by lifting the map $$\mathbb R \mapsto \mathbb S^1 \mapsto M$$, then your interpretation does make sense and is correct.

In your third bullet, the meaning of "properly embedded" is that the parameterization $$\mathbb R \mapsto \widetilde M$$ of $$l$$ is a proper function, meaning one for which the inverse image of every compact subset of $$\widetilde M$$ is a compact subset of $$\mathbb R$$. Intuitively this means that as points in $$\mathbb R$$ go to $$-\infty$$ or to $$+\infty$$, their images in $$\widetilde M$$ also "go to infinity".

In your fourth bullet, your guess is correct, if you restate it slightly: the full preimage will be the union of the $$\pi_1(M)$$ translates of $$l$$.

• Thank you for your answer. Just for clarification: So, in the first bullet you are using covering is a local isometry. In the second and third bullets, a parameterization $\Bbb R\to \widetilde M$ of $l$ is simple as $f_\alpha$ is simple, and that is why the text says $l$ is embedded in $\widetilde M$. Am I right? Jan 18, 2021 at 4:07
• That sounds right Jan 18, 2021 at 4:29
• except that showing $l$ is embedded in $\widetilde M$ does make use of the fact that $\alpha$ is of infinite order. Jan 18, 2021 at 4:30
• Can you elaborate on your last comment whenever you have time? or any reference will also be helpful. Jan 18, 2021 at 5:01
• I think it follows from the lemma 5.2.1(as stated above) as the image of $l$ in $M_\alpha$ that is $\text{im}(f_\alpha)$, represents a generator of $\pi_1(M_\alpha)$. Am I right? Jan 18, 2021 at 5:07