Isomorphism canonical and Moebius bundle.

Question

I have to prove that the canonical bundle over $\mathbb{R}P^1$ is isomorphic to Moebius bundle. We define the caninical bundle as $\xi=\{E^{\perp},p,S^1 \}$ where $E^\perp:=\{(l,v) \in \mathbb{R}P^{1} \times \mathbb{R}^2 \, | \ v \perp l \}$ and $p(l,v)=v$ and the Moebius bundle $\eta:=\{E,q,S^1\}$ where $E:=\frac{[0,1] \times \mathbb{R}}{(0,x)\sim(1,-x)}$. Now we have clearly that Moebius bundle isn't trivial. On the other hand, $\xi$ is isomorphic to the tangent bundle $TS^1$ which is trivial because $S^1$ is a Lie group. Could you please explain to me where do I made a mistake?

Answer 1

You're presumably thinking of $S^1 \subset \mathbf{R}^2$ in the usual way, in which case $E^\perp$ is indeed the total space of $TS^1$. But this $S^1$ is not "naturally" $\mathbf{RP}^1$; it's the connected double cover.

Consequently, the bundle $\xi$ is the quotient of $TS^1$ by the antipodal involution $(x, v) \in S^1 \times \mathbf{R} \mapsto (-x, -v)$, which is the Möbius bundle.