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I'm asked to compute all the possible degrees of a covering space $S^{2n} \to X$, where $X$ is a path connected space. My idea is to try to show that these degrees can only be $1$ (take the identity and $S^{2n}$ as a path connected space) and $2$ (for example the double cover over $RP^{2n}$).

My attempt: My idea is to use the fact that the only non trivial group that act freely on $S^{2n}$ is $\mathbb{Z}_2$. At least, this is the only thing that relate in some way covering, the numbers $1,2$ and even-sphere. So I tried to discover if a covering map induces some kind of actions over $S^{2n}$, and I recall the existence of the deck transformations. I must admit that my knowledge about them are pretty limited, so I'm not feeling so confident with them. My goal would be to prove that this group act (so induces homeo from the sphere to itself) freely. And then finding a way to relate the fact that this group would be $\mathbb{Z}_2$ and the possible degree (=: number of sheets) of the cover.

Is this approach correct? if not, would you mind suggesting me a new starting point to prove or disprove my assumption?

Emilio Ferrucci
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Luigi M
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  • Right, $\pi_1(X)$ always acts freely by deck transformations on the universal cover of $X$. Since $S^{2n}$ is simply connected for $n \geq 1$ it is the universal cover of $X$, so your approach works :) – Emilio Ferrucci Oct 29 '14 at 21:50
  • Thanks @EmilioFerrucci, my problem is my lack of proper knowledge about the deck transformations, so I try to catch up with this kind of things – Luigi M Oct 29 '14 at 22:00
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    @EmilioFerrucci are you studying in Bonn? Me too, first year Msc – Riccardo Oct 29 '14 at 23:42
  • @Riccardo Yes I'm a 2nd year master student. I've probably seen you around! Friend me on fb ;) – Emilio Ferrucci Oct 30 '14 at 13:00

2 Answers2

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If you already know that $\mathbb Z_2$ is the only group that can freely act on $S^{2n}$ then you are done. Just like you said, the group of deck transformations acts freely on a covering space. Also since $S^{2n}$ is simply connected, the group of deck transformations is isomorphic to $\pi_1(X)$. (See for example Prop 1.39 of Hatcher's book Algebraic Topology.) So that means $\pi_1(X)$ is either trivial or $\mathbb Z_2$, corresponding to either a degree $1$ or degree $2$ cover.

Cheerful Parsnip
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This is definitely an addendum to the answer of Grumpy Parsnip (which deserve to be the main answer), but I think it can be useful to be stated as an answer and not as a comment, because the OP explicitly asked for some guidance in the topic.

Recall the notion of normal covering: A covering $$ p \colon \tilde{X} \to X$$ is called normal is for every $x \in X$, and for all $\tilde{x_1},\tilde{x_2}$ in the fibre of a fixed $x$, then there exist a deck transformation which sends $\tilde{x_1} \mapsto \tilde{x_2}$. Proposition $1.39$ of Hatcher's Algebraic Topology tells you that universal covering implies normal covering (the trivial subgroup is trivially normal :) ).

So once you have that $\Delta$ (the group of deck transformations) is either a group with one or two elements, you get that the cardinality of a sheet (which is constant due to path-connectedness of $X$) is either $1$ or $2$.

Riccardo
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