I am studying on my own Lens spaces from the algebraic topology viewpoint. I read about them on Hatcher's book as you can deduce from some of my previous questions:



I have been thinking...is it possible to construct even dimensional Lens spaces?

I have looked in Google but I haven't found anything. So I suspect that maybe the action doesn't work fine on even dimensional spheres. Any reference to look for it? or any argument in one or the other direction?

P:D: Just to remember, this is the definition of Lens space that I am using:


Thanks in advance!


1 Answer 1


For even dimensional spheres, we have the following theorem.

Suppose: Suppose $\pi:S^{2n}\rightarrow L$ is a covering map. Then either $\pi$ is a diffeomorphism or is a $2$-fold covering and $L$ is non-orientable. In particular, $L$ has fundamental group either trivial or $\mathbb{Z}/2\mathbb{Z}$.

Proof: Let $G = \pi_1(L)$ and pick $g\in G$. Consider the induced action of $G$ on $S^{2n}$ as the Deck group.

Then $g:S^{2n}\rightarrow S^{2n}$ is a homeomorphism. Assume initially that this map is orientation preserving. Then it has degree $1$ and thus, according to this MSE question, because $g$ is not homotopic to the antipodal map (because $2n$ is even), $g$ must have a fixed point. But an element of the Deck group which fixes a point must be the identity, so $g = e\in G$.

This shows that the identity is the only orientation preserving element of $G$. What about orientation reversing elements? Well, if $h_1,h_2\in G$ both reverse orientation, then $h_1h_2^{-1}$ preserves orientation, so $h_1h_2^{-1} = e$, so $h_1 = h_2$. Thus, if there is an orientation reversing element of $G$, there is only one.

Thus, $G$ consists of $1$ or $2$ elements, so $G = \{e\}$ or $G=\mathbb{Z}/2\mathbb{Z}$.


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