3
$\begingroup$

I know that this question has already been asked, but I couldn't find a clear answer.

I have to show that the cylinder and the Möbius strip are fiber bundles over $S^1$ with fiber an open interval and I have to write down explicitly the transition functions and the cocycles.

Cylinder.

The cylinder is $E:=I\times S^1$ where $I$ is our open interval.

Let's consider $$ \pi : I \times S^1 \to S^1 $$ the projection on the second factor (by definition it's continuous and surjective).

Let's take $S^1$ as basis of the fiber bundle and let $\{U_\alpha\}_\alpha$ be an open covering of $S^1$.

$\pi_1:U_\alpha\times I\to U_\alpha$ is the projection on the first factor.

The trivialization is $$ \phi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times I\\ \phi_\alpha:=id|_{U_\alpha\times I} $$ In this case the transition functions are such that for all $\alpha,\beta$ $$ \phi_{\alpha\beta}=\phi_\alpha\circ\phi_\beta^{-1}=id|_{(U_\alpha\cap U_\beta)\times I}. $$ Hence the cocycles are just $$ g_{\alpha\beta}:U_\alpha\cap U_\beta\to Aut(I)\\ p\mapsto id|_I $$

Möbius band Let's start with the closed interval $[0,1]$ and the map $$ p:[0,1]\to S^1\\ x\mapsto e^{i2\pi x} $$ which identifies the point $0$ with the point $1$.

Let's consider $[0,1]\times I$, where $I$ is an open interval, let's say $(0,1)$. Let's define an equivalence relation on $[0,1]\times I$: $$ (0,y)\sim (1,1-y) $$ and we obtain the Möbius band $E$.

Let's consider an open covering $\{U_\alpha\}_\alpha$ of sphere $S^1$ and let $$ \pi:E\to S^1\\ [(x,y)]\mapsto p(x) $$ be continuous and surjective and $$ \pi_1:U_\alpha\times I\to U_\alpha $$ be continuous and surjective.

I can't find a trivialization. I just know that I have to choose a proper open covering of $S^1$ like two open interval overlapped.

$\endgroup$
  • 1
    $\begingroup$ Can you find the trivialization in the case that $U_\alpha \subset p(0,1)$? $\endgroup$ – Lee Mosher Jul 29 '15 at 13:52
  • 1
    $\begingroup$ Yes! Let's define $U_1:=S^1-\{0\}=p((0,1))$. Then I want to construct a homeomorphism $\phi_1:\pi^{-1}(U_1)\to U_1\times I$. Let's see what $\pi^{-1}(U_1)$ is. $\pi^{-1}(U_1)=\{[(x,y)]\in E: p(x)\in p((0,1))\}=(0,1)\times I$. Hence $\phi_1:\pi^{-1}(U_1)\to U_1\times I$ maps $[(x,y)]$ to $(p(x),y)$. Right? $\endgroup$ – avati91 Jul 29 '15 at 14:16
  • $\begingroup$ Yup. That knocks off one big case. Now all you have to do is figure out the complementary case $U_\alpha \not\subset p(0,1)$. $\endgroup$ – Lee Mosher Jul 29 '15 at 17:15
  • $\begingroup$ I think I solved my doubts :) Thanks $\endgroup$ – avati91 Jul 29 '15 at 17:20
  • 2
    $\begingroup$ The example of gluing a Mobius strip from a cocycle is covered in Jeffrey Lee's book in Example 6.16 in case you want a textbook source. $\endgroup$ – ಠ_ಠ Oct 29 '15 at 4:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.