Are two vector bundles Möbius band and $S^1\times \mathbb{R}$ isomorphic as vector bundles over $S^1?$

Question

If $\forall (x,t), (y,t^{\prime}) \in \mathbb{R}\times \mathbb{R}$ define $$(x,t)\backsimeq(y,t^{\prime})\Leftrightarrow \exists n\in \mathbb{Z} ; x=y+2n\pi , t=(-1)^{n} t^{\prime} $$ $$x\backsim y \Leftrightarrow \exists n\in \mathbb{Z} ; x=y+2n\pi$$ Then $S^{1}= \mathbb{R}/ \backsim$ and $E:= \mathbb{R}\times \mathbb{R}/\backsimeq$ is called Möbius band.

Question: Are two vector bundles Möbius band and $S^1\times \mathbb{R}$ isomorphic as vector bundles over $S^1?$ (Isomorphism bundle)

Both have the same space and the same standard fiber; but $S^1\times \mathbb{R}$ is orientable, while the other is not.

Question2: What about vector bundles $HTM$ and $\pi^*T^*M$ over TM?

Here $TTM= HTM \oplus VTM$.

Fiber over $u\in TM$ is as the following $$H_{u}TM:= \{\alpha\in T_{u}TM | \pi_{*}(\alpha)= u\}\to u \in TM,$$ $$\pi^*T^*M\subset TM\times T^*M\to TM\quad, \quad \pi^*T^*M := \{(v, \alpha)| \pi(v)=\tau(\alpha)\}\to v,$$ where $$\pi:TM\to M \quad; \quad (x,y)\to x,$$ and $$\tau:T^*M \to M \quad , \quad (x, \alpha)\to x.$$ How can I prove $HTM$ and $\pi^*T^*M$ over $TM$ are isomorphic? How do I guess bundle isomorphism between them?

Thanks for any hint.

Answer 1

First question: no, of course!

First motivation: the Möbius strip $M$ is not orientable and the cylinder $\mathbb{S}^1\times\mathbb{R}=C$ is orientable; if $M$ and $C$ are isomorphic as vector bundles then $M$ and $C$ are diffeomorphic as manifolds, this is a contraddiction; because $M$ is not orientable and $C$ is orientable.

Second motivation: because $C$ and $M$ are topological subspaces of $\mathbb{R}^3$, $C$ is a closed non compact subset of $\mathbb{R}^3$ and $\overline{M}$ (the closure of $M$) is a compact subset of $\mathbb{R}^3$; then $C$ and $M$ are not homeomorphic subspaces of $\mathbb{R}^3$, in particular they can not be diffeomorphic manifolds, and therefore they are not isomorphic line bundles over $\mathbb{S}^1$.

Second question: either you define an Ehresmann connection on $TM$, so you can determine a horizontal subbundle $H(TM)$ of $T(TM)$, or you specify which is the subbundle $H(TM)$ of $T(TM)$!