I How Do Transition Functions Define a Möbius Strip in Fiber Bundles?

[cianfa72]

TL;DR Summary

Möbius strip fiber bundle defined through 'gluing instructions' between its two charts

Hi,

starting from this (old) thread

But suppose instead that the $ϕ_{12}(b)$ is the identity map on $F$ for all $b$ in $X$, and the map $f↦−f$ for all $b$ in $Y$. Then the bundle is a Möbius strip.

I'm a bit confused about the following: the transition function $ϕ_{12}(b)$ is defined just on the intersection $U_1\cap U_2$ and as said in that thread it actually amounts to the 'instructions' to glue together the two charts to obtain the Möbius strip.

Neverthless my doubt is: what about the 2 points in $S^1$ (the base space) not taken in account ? On that points (and for the corresponding fibers) we actually do not define any 'gluing instruction' ?

 

[lavinia]

The gluing instruction is to paste with a reflection, multiplication by $-1$ if you view the fibers as copies of the interval $[-1,1]$.

If you use circles rather than intervals as the fibers then the gluing instruction to paste by reflection around an axis of the circle gives a fiber bundle whose fibers are circles. Geometrically it is a closed surface called the Klein Bottle.

 

[cianfa72]

The gluing instruction is to paste with a reflection, multiplication by $-1$ if you view the fibers as copies of the interval $[-1,1]$.

Sorry, as far as I understood, the intersection $U_1\cap U_2$ is made of two disjoint subsets (name it $X$ and $Y$).

For Möbius strip the gluing instruction for fibers 'staying' onto base points belonging to $A$ are different from that 'staying' on point belonging on $B$. Namely for fibers staying on $A$ points the gluing instruction is "paste without reflection" rather than with reflection (multiplication by -1) for fibers staying on $B$ points

 

[lavinia]

Sorry, as far as I understood, the intersection $U_1\cap U_2$ is made of two disjoint subsets (name it $X$ and $Y$).

For Möbius strip the gluing instruction for fibers 'staying' onto base points belonging to $A$ are different from that 'staying' on point belonging on $B$. Namely for fibers staying on $A$ points the gluing instruction is "paste without reflection" rather than with reflection (multiplication by -1) for fibers staying on $B$ points

On one interval paste without reflection. On the other paste with reflection. Imagine making a Mobius band out of a strip of paper and it will be clear.

 

[cianfa72]

On one interval paste without reflection. On the other paste with reflection. Imagine making a Mobius band and it will be clear.

That's clear to me. My concern is about the two points that do not belong to the intersection $U_1\cap U_2$ thus to neither $X$ nor $Y$. For the fibers 'over' them (preimage under bundle projection of those two points in the base space) the gluing procedure does not apply at all ?

 

[lavinia]

That's clear to me. My concern is about the two points that do not belong to the intersection $U_1\cap U_2$ thus to neither $X$ nor $Y$. For the fibers 'over' them (preimage under bundle projection of those two points in the base space) the gluing procedure does not apply at all ?

I am not sure what two points you mean.

The picture I have is the circle covered by two over lapping segments that intersect in two disjoint open intervals.

On one interval the transition function is the identity and on the other it is reflection.

 

[cianfa72]

The picture I have is the circle covered by two over lapping segments that intersect in two disjoint open intervals.

These two disjoint open intervals do not cover the entire circle $S^1$ , right ?
So, what about fibers over the other (two) points belonging to $S^1$ ? That is my question !

 

[lavinia]

These two disjoint open intervals do not cover the entire circle $S^1$ , right ?
So, what about fibers over the other (two) points belonging to $S^1$ ? That is my question !

The rest of the circle you do nothing. There is no transition function.

 

[cianfa72]

The rest of the circle you do nothing. There is no transition function.

As explained by @Fredrik in that thread, take the base $B=S^1$ and let $U_1=S^1-\{(0,1)\}$ and $U_2=S^1-\{(0,-1)\}$.

Now $U_1 \cap U_2$ --made of the two disjoint subsets $X$ and $Y$ of $S^1$-- does not cover enterely $S^1$ because $\{(0,1)\}$ and $\{(0,-1)\}$ are not included there.

 

[lavinia]

As explained by @Fredrik in that thread, take the base $B=S^1$ and let $U_1=S^1-\{(0,1)\}$ and $U_2=S^1-\{(0,-1)\}$.

Now $U_1 \cap U_2$ --made of the two disjoint subsets $X$ and $Y$ of $S^1$-- does not cover enterely $S^1$ because $\{(0,1)\}$ and $\{(0,-1)\}$ are not included there.

Right.

Gluing happens only on intersections.

 

[zinq]

I would suggest that, when quoting a problem to start a new thread, quote the whole problem.