Covering $\Bbb RP^\text{odd}\longrightarrow X$, what can be said about $X$?

Question

I am looking for any argument related to the following fact, which may or may not be true.

Let $f:\Bbb RP^n\longrightarrow X$ be a covering space, where $n\geq 2$. Then, $X=\Bbb RP^n$.

Now, for $n=\text{even}$, this is surely true and this is given below :

Let $f:\Bbb RP^n\to X$ be a covering space, where $n=2m$ for some $m\in\Bbb N$. Then, $X$ is compact connected $2m$-manifold. So, $X$ is a finite CW-complex. Also, $f$ is a finite sheeted covering as fibres are discrete subsets of the compact space $\Bbb RP^n$. Let $f$ be $k$-sheeted covering. Then, $$1=\chi(\Bbb RP^n)=k\cdot \chi(X)\implies k=1=\chi(X).$$ Now, single-fold covering is a homeomorphim, so we are done.

So, my question is what about $n=\text{odd}$, here $\chi(\Bbb RP^{\text{odd}})=0$, probably we can not modify the above argument. Is there any alternative argument to prove above? Is there any $X$ not homeomorphic to $\Bbb RP^\text{odd}$ with a covering $\Bbb RP^\text{odd}\longrightarrow X$.

Answer 1

If $n = 1$, then $\mathbb{RP}^1 = S^1$ which only covers itself.

If $n > 1$, the manifold $\mathbb{RP}^{2n-1}$ covers infinitely many manifolds which are pairwise non-homotopy equivalent. To see this, we use the fact that $\mathbb{RP}^{2n-1}$ is a lens space.

Recall, if we identify $S^{2n-1}$ with the unit vectors in $\mathbb{C}^n$, then for positive integers $m, l_1, \dots, l_n$ with $(m, l_i) = 1$, the lens space $L(m; l_1, \dots, l_n)$ is the quotient of $S^{2n-1}$ by $\mathbb{Z}_m$, where the action of $\mathbb{Z}_m$ is generated by $(z_1, \dots, z_n) \mapsto (e^{2\pi i l_1/m}z_1, \dots, e^{2\pi i l_n/m}z_n)$. In particular, when $m = 2$ and $l_1 = \dots = l_n = 1$, the $\mathbb{Z}_2$ action on $S^{2n-1}$ is given by $(z_1, \dots, z_n) \mapsto (-z_1, \dots, -z_n)$ which is the antipodal map, and hence $L(2; 1, \dots, 1) = \mathbb{RP}^{2n-1}$.

For $k > 0$, consider the lens space $L(2k; 1, \dots, 1)$. The $\mathbb{Z}_{2k}$-action on $S^{2n-1}$ has an index $k$ subgroup which is just the antipodal action: if $g$ is a generator of the $\mathbb{Z}_{2k}$-action, then $g^k$ is the antipodal map and $\{\operatorname{id}, g^k\}$ is the desired subgroup. It follows that $L(2k; 1, \dots, 1)$ has $L(2; 1, \dots, 1) = \mathbb{RP}^{2n-1}$ as a $k$-sheeted cover.

The manifolds $L(2k; 1, \dots, 1)$ are homotopically distinct for every $k$ because $\pi_1(L(2k; 1, \dots, 1)) \cong \mathbb{Z}_{2k}$.