$\Bbb{R}P^1$ bundle isomorphic to the Moebius bundle

Question

I'm trying to construct an explicit isomorphism from $E = \{([x], v) : [x] ∈ \Bbb{R}P^1, v ∈ [x]\}$ to $T = [0, 1] × R/ ∼$ where $(0, t) ∼ (1, −t)$. I verified that $\Bbb{R}P^1$ is homeomorphic to $\Bbb{S}^1$ which is homeomorphic to $[0,1]/∼$ where $0∼1$. So this is the map I have in my mind: $([x],v)\to (x,(1-x)v+xe^v)$. Does that work? It doesn't look very natural.

Answer 1

In your formula, it looks as if "$x$" denotes both a non-zero vector in the plane (since you write $[x] \in \mathbf{RP}^1$) and a parameter in a convex linear combination...?

To see and construct your bundle isomorphism explicitly, it may help to lift your maps to the connected double covers $\widehat{E}$ of $E$ and $\widehat{T}$ of $T$.

The unit circle in the plane double covers $\mathbf{RP}^1$, and the fibre involution $t \mapsto -t$ "unwraps" to the identity map. Thus, you may view $\widehat{E}$ as the cylinder $S^1 \times \mathbf{R}$. (For notational convenience, take $S^1 \subset \mathbf{C}$ below.)

Taking $\widehat{T} = \bigl([-1, 1] \times \mathbf{R}\bigr)/\sim$ with $(-1, t) \sim (1, t)$, the parametrization $(\theta, t) \in \widehat{T} \mapsto (e^{i\theta}, t) \in \widehat{E}$ defines a bundle isomorphism between the double covers.

The double-covering map $\pi:\widehat{E} \to E$ is the quotient by the "antipodal" map $\pi(x, t) = (-x, -t)$, and the double-covering map $\widehat{T} \to T$ is the quotient by $\varpi(x, t) = \bigl(x + 1 \pmod{2}, -t\bigr)$. From here it should be straightforward to find your desired mapping from $T$ to $E$. (It's formally easier to write down this direction, since you have a fairly obvious mapping $[0, 1] \times \mathbf{R} \to E$, and only need to check the map is constant on equivalence classes, i.e., factors through the quotient to $T$.)

In case these additional observations are helpful: $S^1 \times \mathbf{R}$ is diffeomorphically viewed as the punctured plane via the "polar coordinate mapping" $(e^{i\theta}, t) \mapsto e^{t + i\theta} = re^{i\theta}$ (with $r= e^t$). The ray through the origin and $e^{i\theta}$ may be viewed as representing the line through the same two points. Since $e^{i\theta}$ and $e^{i(\theta + \pi)} = -e^{i\theta}$ determine the same line in the plane (and therefore the same point in $\mathbf{RP}^1$), the punctured plane double-covers $E$. The antipodal map on the cylinder corresponds to inversion in the unit circle followed by a half-turn of the plane, and (if you Really Understand this picture) the quotient is clearly the tautological bundle over $\mathbf{RP}^1$.