# Bundles and global sections, triviality.

1. Mar 16, 2009

### WWGD

Hi:
I am trying to understand more geometrically the relation between triviality

of bundles and existence of global sections. This is what I have for now. Please

comment/critique:

Let p:E-->B be a fiber bundle :

consider E embedded in B as the 0 section. Then , if the bundle is trivial, E

is a (global) product space; E=BxF )so that p(b,f)=b).

Then every continuous map from B to E is a global section:

I am trying to understand why we can define a global section on S^1xI as

a bundle over S^1 , but not in M, the Mobius band , as a bundle over S^1 .

It seems that the "torsion" of M (I think it is measured in Chern Classes, or

Characteristic classes. ) prevents this from happening. Anyone know, at

least intuitively how the twisting prevents a global section?.

My friend told me that we take the Mobius band as an identification/quotient space

( a square, with sides given orientations , and sides identified), say

S=IxI , and we consider the strip {1/2xI}, as the base, embedded in S .

Then, if we have a continuous map f from {1/2xI} into I , which is nowhere-zero, the

twisting (when identifying the sides of S with opposite orientation) will force f to

take on both negative and positive values . This seems intuitively correct, but too

fuzzy. How do we rigorize this?.

Then, by continuity , f must be zero at some point.

Thanks.

2. Mar 16, 2009

### yyat

To talk about the 0-section you need a vector bundle. For a general fiber bundle the 0-section is not defined. In fact, a fiber bundle can have no global sections at all.

You mean from B to F.
S^1xI and the Möbius band are of course not vector bundles, but one can take S^1xR and the Möbius bundle instead (IxR with opposite sides glued in reverse orientation), which is homeomorphic to the Möbius band without the boundary.
Also, you probably mean a nonvanishing global section. The Möbius bundle has many global sections, but they all vanish at some point.
A vector bundle has a global section iff the Euler class vanishes.

To see that the Möbius bundle can not be isomorphic to S^1xR (this is equivalent to the existence of a global nonvanishing section), note that the complement of the zero section in S^1xR has two connected components, but the complement of the zero section in the Möbius bundle has only one. You can verify this without too much difficulty from the definition, or by constructing a paper model and cutting it along the middle.

3. Mar 16, 2009

### WWGD

Yes, sorry for my carelessness. Maybe I should not post at 2a.m after a long day.

Anyway: what are independent sections, as in a vector bundle (R^n bundle) is

trivial if it has n independent sections?. I understand a section is a map from

the base to the top space in which every point maps to its fibre. If

these sections are, say, {f_1,..,f_n} , and f_i: B-->R^n , do we mean that

the vectors {f_i(b)} are linearly independent?

Thanks.

4. Mar 16, 2009

### yyat

Yes, linear independent sections are just pointwise linearly independent (by definition).

5. Mar 20, 2009

### wofsy

The Mobius band can be thought of as a rectangle with two of the opposite edges identified by a reflection.

to be concrete take the rectangle [0,1] x [-1,1] and identify the two vertical edges by (0,y) - > (1,-y)

A section would be a map from the unit interval on the x-axis into this rectangle x -> (x,S(x)) with the property that S(0) is the negative of S(1). By the intermediate value theorem this can not happen unless S crosses zero.

Generally if you have n independent sections of a n-plane bundle then there is a map from BxR^n into the total space that maps (x,a1,...,an) to a1.S1(x) + ....+ an.Sn(x). It is easy to show that this map is a homeomorphism

6. Mar 20, 2009

### WWGD

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7. Mar 20, 2009