# Fundamental group via deck transformations considering rotation on sphere

Let $$Z_m$$ act on $$S^1$$ by multiplication with $$e^{2\pi ki/m}$$ for $$k \in Z_m$$. Let $$X = S^1 / Z_m$$ be the orbit space of this action. Then we have a universal cover $$q:S^1 \rightarrow X$$ given by the canonical projection. To determine the fundamental group of $$X$$ we can consider the group of deck transformations $$Aut(q)$$ of $$q$$, since we know that $$\Pi_1(X) = Aut(q)$$.

$$Aut(q)$$ is by definition the group of all homeomorphisms $$f: S^1 \rightarrow S^1$$ such that $$qf=q$$. We see that all rotations by degree $$e^{2\pi ki/m}$$ for $$k \in Z_m$$ are such homeos. But why are there no other ones?

• Since $S^1$ is not simply connected, your map $q$ is not a universal cover of $X$, so we can't say $\pi_1 (X) \cong \text{Aut}(q)$. In fact, it is correct that $\text{Aut}(q) \cong \mathbb{Z}/m\mathbb{Z}$, but $\pi_1 (X) \cong \mathbb{Z}$, since $X$ is homeomorphic to $S^1$. – Sameer Kailasa Jan 2 at 19:43
• yes you're right I have been actually thinking about lens spaces which come from a group action on $S^3$ but I thought I could make things easier for my question by considering only $S^1$ – CHwC Jan 2 at 19:52
• I want to kind of understand why we cannot find other homeos of $S^1$ which permute the orbits than the rotations – CHwC Jan 2 at 19:53

$$\pi_1(X)$$ is not meant to be equal to $${\rm Aut}(q)$$. That would be true if the covering was a universal cover, which is not the case here.

[An example of a universal cover of $$X$$ would be the map $$q_{\rm univ} : \mathbb R \to X$$, sending $$x \mapsto x {\rm \ mod \ } \tfrac{2\pi}{m}$$. This is a universal cover because $$\mathbb R$$ is simply connected. The deck transformation group for this universal cover is $${\rm Aut}(q_{\rm univ}) = \mathbb Z$$. Since $$X$$ itself is a small circle, $$\pi_1(X) = \mathbb Z$$, which agrees with $${\rm Aut}(q_{\rm univ})$$.]

As for your original covering $$q : S^1 \to X$$, we still have the following useful result (Hatcher 1.39): If $$H : = q_\star (\pi_1(S^1))$$ is the image of $$\pi_1(S^1)$$ under the group homomorphism $$q_\star : \pi_1(S^1) \to \pi_1(X)$$, then $${\rm Aut}(q)$$ is isomorphic to $$N(H) / H$$, where $$N(H)$$ is the normaliser of $$q_\star (\pi_1(S^1))$$ in $$\pi_1(X)$$. Now $$\pi_1(X) = \mathbb Z$$, and $$H$$ is the subgroup $$m\mathbb Z$$, which is a normal subgroup, so the theorem tells us that $${\rm Aut}(q) = \mathbb Z_m$$. And this agrees perfectly with your counting.

Finally, I'll address the question in your comment: Can we see directly that there are no more than $$m$$ deck transformations for $$q$$? The answer is yes, because $$q$$ is an $$m$$-sheeted covering of $$X$$ (meaning that the preimage of $$q^{-1}(x)$$ for any point $$x \in X$$ consists of $$m$$ points. Given a point $$x \in X$$, a deck transformation $$f$$ is uniquely determined by where it sends $$x$$. But $$f$$ can only send $$x$$ to one of the points in $$q^{-1}(x)$$, and there are $$m$$ such points. Hence there can be at most $$m$$ deck transformations.

• okay but is there any way to see "directly" why there can be no other homeos which satisfy $qf=q$ then the rotations? – CHwC Jan 2 at 19:50
• when I try to imagine such other homeos then I have to twist things around so much that I feel like the continuity of these maps is lost somewhere, but I dont see how to express this formally – CHwC Jan 2 at 19:51
• @CHwC See edit... – Kenny Wong Jan 2 at 20:02
• why is a deck transformation uniquely determined by the value it takes on an arbitrary point x? i guess thats the point that i am missing – CHwC Jan 2 at 21:35
• @CHwC This is a consequence of the Unique Lifting Property (Hatcher 1.34): Given a covering space $q : \widetilde X \to X$ and a map $f : Y \to X$ with two lifts $\widetilde f_1, \widetilde f_2 : Y \to \widetilde X$ that agree at one point of $Y$, then if $Y$ is connected, these two lifts must agree on all of $Y$. If you apply this with $Y = \widetilde X$ and $f = q$, then $\widetilde f_1, \widetilde f_2$ are deck transformations, so the theorem tells you that if two deck transformations agree at one point, then they agree everywhere. – Kenny Wong Jan 2 at 21:40