Must a Smooth Section Over a Mobius Strip Take Value Zero at Some Point?

[cianfa72]

TL;DR Summary

About smooth sections over the Mobius strip

As discussed in a recent thread, I'd ask whether any smooth section over a Mobius strip must necessarily take value zero on some point over the base space $\mathbb S^1$.

Edit: my doubt is that any closed curve going in circle two times around the strip is not actually a section at all.

Thanks.

 

[jbergman]

TL;DR Summary: About smooth sections over the Mobius strip

As discussed in a recent thread, I'd ask whether any smooth section over a Mobius strip must necessarily take value zero on some point over the base space $\mathbb S^1$.

Edit: my doubt is that any closed curve going in circle two times around the strip is not actually a section at all.

Thanks.

Remember the square quotienting opposite sides with a flip I described before.

Imagine you draw a line across the square starting at (0, 1). This line has to end at (1,0) to be continuous as those two points are identified. How can you draw a line from (0,1) to (1,0) without crossing the line y=1/2 (the zero line in this description)?

 

[cianfa72]

Imagine you draw a line across the square starting at (0, 1). This line has to end at (1,0) to be continuous as those two points are identified. How can you draw a line from (0,1) to (1,0) without crossing the line y=1/2 (the zero line in this description)?

Yes, to be continuous that line must cross the y=1/2 line necessarily.

Regarding a continuous line that goes around two times around the strip, I think it is not a section since, by definition, a section $\sigma$ is a continuous map from the base space $\mathbb B$ into the bundle $\mathbb E$ (i.e. every point in the base space has exactly one image under $\sigma$).

 

[jbergman]

Yes, to be continuous that line must cross the y=1/2 line necessarily.

Regarding a continuous line that goes around two times around the strip, I think it is not a section since, by definition, a section $\sigma$ is a continuous map from the base space $\mathbb B$ into the bundle $\mathbb E$ (i.e. every point in the base space has exactly one image under $\sigma$).

You are correct. But when we were talking about foliating the Mobius strip we are thinking about foliating the total space of the vector bundle or alternatively one can just define a smooth structure on the quotiented square directly and foliate that as I described.

 

[cianfa72]

But when we were talking about foliating the Mobius strip we are thinking about foliating the total space of the vector bundle or alternatively one can just define a smooth structure on the quotiented square directly and foliate that as I described.

I believe you're talking about different ways to look at the definition of Mobius strip's differential structure.

In the former one defines "intrinsecally" the Mobius strip as the quotiented square with two overlapped charts. In the latter, instead, one uses its embedding in $\mathbb R^3$.

Basically the above two definitions are actually equivalent.