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

[cianfa72]

TL;DR

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.

 

[lavinia]

A more general way to look at this is to observe(proof?) that a line bundle with a non-zero section must be trivial. This is true for any line bundle over any space.

As a line bundle over the circle, the Mobius strip is not trivial. If it were it would be homeomorphic to a cylinder.

 

[jbergman]

A more general way to look at this is to observe(proof?) that a line bundle with a non-zero section must be trivial. This is true for any line bundle over any space.

As a line bundle over the circle, the Mobius strip is not trivial. If it were it would be homeomorphic to a cylinder.

Your section is a global frame and the existence of a frame on an open set U is equivalent to a trivialization.

 

[lavinia]

A couple of related ideas:

The parity of the number of zeros of a transversal section of the Mobius strip bundle

Any smooth section of the Mobius strip, viewed as a line bundle over the circle, that has finitely many zeros and is transversal* to the zero section must have an odd number of zeros. If the number were even the bundle would be trivial. (proof?) and the total space of the bundle would be a cylinder not a Mobius strip. The parity of the number of zeros tells you whether the bundle has a non-zero section.

* Two submanifolds of a manifold,M, intersect transversally if at each intersection point their tangent spaces span the entire tangent space to M. In the case of the Mobius band the two submanifolds are the base circle and the section. Their tangent spaces at intersection points intersect only at the origin and span the two dimensional tangent space of the Mobius band at the point of intersection.

Circles on the Mobius strip viewed as sphere bundles over the circle

The circles that are parallel to the equator of the Mobius strip are twice as long as the equator and cover the equator twice when projected onto it. Two points on these circles lie above each point on the equator.

From the point of view of line bundles this is no surprise since in each fiber one can choose two points that are the same distance to the equator. In general, on any vector bundle with a Riemannian metric, points in each fiber that are a fixed distance to the origin of the fiber form a sphere and this creates a new fiber bundle, not a vector bundle, called a sphere bundle. A parallel circle of the Mobius strip is a 0 dimensional sphere bundle over the base circle.

An equivalent way to see that the Mobius strip bundle is non trivial - equivalent to showing that every section must have a zero - is to notice that the sphere bundle is connected. It is a single circle. If the bundle were trivial the sphere bundle would be split into two circles.

Finally it should be no surprise that the Mobius strip with its equator cut out is also connected (since each parallel circle is connected) and is not separated into two pieces as is a cylindrical strip.

Prove that the Mobius band minus its equator is the Cartesian product of a circle with a line segment. That is: it is homeomorphic to a cylinder. This means that a second cut along a circle will separate the strip into two pieces.

The Tangent Bundle to the Mobius strip

The tangent bundle to the Mobius strips splits into the direct sum of two lines bundles, one the lines tangent to the equator and the parallel circles to it, call it E, and the other tangent to the fiber lines of the Mobius line bundle, call it F.

A few things to think about:

The line bundle, E, is trivial.
The line. bundle ,F, restricted to the equator is just the Mobius bundle back again.
The line bundle,F, away from the equator is trivial.
If one identifies the boundary circle of the Mobius strip to itself to make a Klein bottle, show that these two line bundles extend to the Klein bottle. So the tangent bundle to the Klein bottle is the direct sum of two line bundles one of which is non-trivial. How does the non-trivial direct summand imply that the Klein bottle is not orientable?

 

[lavinia]

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.

Actually a continuous section must have a zero as well. This is just the intermediate value theorem.