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I want to show that $E=X/{\sim}$ (where $X=[0,1]\times\mathbb{R}^n$) is the total space of a smooth vector bundle over $S^1$. Here $\sim$ is defined as follows: fix a linear isomorphism $L$ of $\mathbb{R}^n$ and set $(0,v)\sim(1,Lv)$ for all $v\in\mathbb{R}^n$.

Here's what I've thought so far and my idea for solving this.

There is a smooth map $\pi:E\longrightarrow S^1$ that maps points $[(t,v)]$, $0<t<1$, to points $s(t)\in S^1$, $s(t)\neq \mathbf{1}$, where $\mathbf{1}\in S^1$ is the point where the gluing of $[0,1]$ took place ($s(t)$ is $p(t)$ where $p:[0,1]\longrightarrow S^1$ is just the defining quotient map of $S^1$), and $\pi$ maps $[(0,v)]=[(1,v)]$ to $\mathbf{1}\in S^1$.

One local trivialization is the easier one, which is where there is no gluing so basically it is the passage to the quotient of the identity map of $(0,1)\times \mathbb{R}^n$. I'm stuck with the other trivialization. My idea is to construct a map $[0,\epsilon)\cup (1-\epsilon,1]$ into itself such that $(t,v)$ is sent to $(t,L_t v)$ where $L_t$ is a family of isomorphisms that "approaches $L$" smoothly as $t\to 0$.

Is this a good idea? If so, could you show me how to fill in the details?

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  • $\begingroup$ Did you mean $Lv$ when you wrote $Tv$? $\endgroup$
    – Thomas Andrews
    Jun 10, 2013 at 17:08
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    $\begingroup$ It may be worth pointing out that your approach won't always work: there may not exist such an $L_t$. More specifically, $L_t$ exists iff $L$ is orientation preserving. $\endgroup$
    – Jason DeVito - on hiatus
    Jun 10, 2013 at 17:12
  • $\begingroup$ @Thomas: yes. I corrected it. Thank you. $\endgroup$
    – Weltschmerz
    Jun 10, 2013 at 17:32
  • $\begingroup$ @Jason: thank you. How does one prove that $L_t$ exists iff $L$ is orientation-preserving? $\endgroup$
    – Weltschmerz
    Jun 10, 2013 at 21:16
  • $\begingroup$ If $L_t$ exists, then the map $[0,1]\rightarrow \mathbb{R}$ given by $t\mapsto \det(L_t)$ cannot have $0$ in the image (since $L_t$ is invertible) and has $1 (= \det(I))$ in the image, hence the image lies in the positive real numbers. For the reverse direction, given $A$ with $\det(A) >0$, one applies the orientation preserving row operations one at a time to show that $A$ can be connected via a path to a diagonal matrix whose diagonal consists solely of $\pm 1$ and further, that there are an even number of $-1$s appearing. Now, one using appropriate rotation matrices to move this to $I$. $\endgroup$
    – Jason DeVito - on hiatus
    Jun 11, 2013 at 1:28

1 Answer 1

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Recall how do you construct the Möbius strip.

Localization is easy, the point is how they (two copies of $I\times \mathbb{R^n}$) glued together on the overlap. (I do not think you need to use $L_t$).

Details: Glue $(0,0.6)\times \mathbb{R}^n$ and $(0.5, 1.1)\times \mathbb{R}^n$ via $(x, v)\mapsto (x, v)$ if $x\in (0.5,0.6)$, and $(x, v)\mapsto (x+1, Lv)$ if $x\in (0, 0.1)$.

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  • $\begingroup$ Thank you. Could you please give me more details? I'm very new in the subject. $\endgroup$
    – Weltschmerz
    Jun 10, 2013 at 17:35

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