$SO(3)$ double covers $L(4,1)$

Question

Let $P^2$ be the real projective plane. I am trying to show that its unit tangent bundle (for a fixed arbitrary metric on $P^2$) is a lens space $L(4,1)$. It seems that this paper (https://www.maths.ed.ac.uk/~v1ranick/papers/konno.pdf) proves a much more stronger result, but I'm seeking for more elementary proof than this. One idea that I am thinking is as follows. The unit tangent bundle of $S^2$ is well-known to be diffeomorphic to $SO(3)$, and it double covers the unit tangent bundle of $P^2$, so maybe if $L(4,1)$ is the only space double covered by $SO(3)=\Bbb RP^3$, then maybe I can get the result. (But I'm not sure about my approach. Is it true that a connected space that is n-fold covered by the 3-sphere must be a lens space $L(n,k)$ for some $k$?) Or is there another possible approach? Thanks in advance.

Answer 1

It's not the case that every $n$-fold covering map $S^3\to M$ must have $M=L(n,k)$, since quotienting $S^3\cong SU(2)$ by nonabelian discrete subgroups will yield covering maps whose codomains are not Lens spaces of any kind. Rather than relying on some uniqueness result, it's possible here to be more explicit. One way is to regard $S^3$ as the set of unit quaternions, and construct the following isomorphisms:

• $S^3/\{\pm 1\}\cong SO(3)$
• $S^3/\{\pm 1,\pm i\}\cong L(4,1)$
• $T_{unit}S^2\cong SO(3)$
• $T_{unit}P^2\cong SO(3)/\{I,R\}$

Where $R\in SO(3)$ is an arbitrary element of order 2. It follows from these that $$ L(4,1)\cong (S^3/\{\pm 1\})/\{[1],[i]\}\cong SO(3)/\{I,R\}\cong T_{unit}P^2 $$