A Pathological example of a Principle Fiber Bundle?

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Geofleur

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In Hassani's Mathematical Physics, a principal fiber bundle is defined as shown below.

principle-fiber-bundle-hassani.jpg

I wanted to see if there is a way to view a tangent bundle as a PFB, even if the resulting structure would have to be globally trivial, so I came up with this idea:

Let ##P = {\rm I\!R} \times {\rm l\!R}##, ##G## be the additive group ##(
{\rm I\!R},+)##, and for ##(a,b) \in P## define the action of ##g## as ##(a,b)g = (a,b+g)##, i.e., ##G## consists of translations of the real line. Then a typical element of ##P/G## is ##[(a,0)] = \{(a,r)|r\in {\rm l\!R}\}##. Let ## \pi((a,b)) = [(a,0)] ##, and let the map ##s_u## be defined by ##s_u((a,b)) = b##. Then we have ##s_u(pg)=s_u((a,b)g)=s_u((a,b+g))=b+g##, while ##s_u(p)g = s_u((a,b))g = bg = b+g##.

Does this structure count as a PFB, albeit one that can only ever be trivial?
 

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lavinia

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In Hassani's Mathematical Physics, a principal fiber bundle is defined as shown below.

View attachment 238350
I wanted to see if there is a way to view a tangent bundle as a PFB, even if the resulting structure would have to be globally trivial, so I came up with this idea:

Let ##P = {\rm I\!R} \times {\rm l\!R}##, ##G## be the additive group ##(
{\rm I\!R},+)##, and for ##(a,b) \in P## define the action of ##g## as ##(a,b)g = (a,b+g)##, i.e., ##G## consists of translations of the real line. Then a typical element of ##P/G## is ##[(a,0)] = \{(a,r)|r\in {\rm l\!R}\}##. Let ## \pi((a,b)) = [(a,0)] ##, and let the map ##s_u## be defined by ##s_u((a,b)) = b##. Then we have ##s_u(pg)=s_u((a,b)g)=s_u((a,b+g))=b+g##, while ##s_u(p)g = s_u((a,b))g = bg = b+g##.

Does this structure count as a PFB, albeit one that can only ever be trivial?
Yes. This also works for the tangent bundle to the circle.

What do you mean by trivial?
 

Geofleur

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By trivial, I mean that ##P## can be identified with the Cartesian product of ##M = P/G## and ##G##. ##G## is just
##{\rm I\!R}## (considered as an additive group) and ## P/G ## can be identified with ##{\rm I\!R}## by identifying ##[(a,0)]## with ##a \in {\rm I\!R}##. Hence, ##P = {\rm I\!R}\times {\rm I\!R} \cong P/G\times G##.

If we look at something like the Mobius strip, it can only be written locally this way, so it's locally trivial, but there is no way to write it globally this way. The best we can do is patch together a bunch of local trivializations using transition functions. That is, for any two fibers ##\pi^{-1}(x_1)## and ##\pi^{-1}(x_2)## over ##x_1,x_2 \in M##, we have a diffeomorphism ##g## from one to the other that must have the same structure as ##G.##

For my proposed example above, each fiber must be a translation of another. But then we can't change the orientation of any of them relative to the others like what would have to be done to form something like the Mobius strip.

I'm guessing that's why the example I came up with seems never to be mentioned in textbooks as an example of a PFB - it's too "boring". But then, there's always the possibility that I'm confused and have gotten something wrong! I like to minimize that probability by posting here :⁠)
 

fresh_42

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'm guessing that's why the example I came up with seems never to be mentioned in textbooks as an example of a PFB - it's too "boring". But then, there's always the possibility that I'm confused and have gotten something wrong! I like to minimize that probability by posting here :⁠)
At least "boring" describes it better than "pathological" :wink:
But I've seen a few things that started out "boring" and then developed a life.
 

lavinia

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- Trivial principal bundles are not pathological. Any space Cartesian product a Lie group is a trivial principal bundle.

There are many manifolds with trivial tangent bundles, for instance every closed orientable 3 manifold and every Lie group. A trivialization of an ##n## dimensional vector bundle determines a principal ##R^{n}## bundle where ##R^{n}## is viewed as a group of translations. Different trivializations determine different principal bundles.
 

Geofleur

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for instance every closed orientable 3 manifold
Does that mean that you can comb a hairy ball if it happens to be the 3-sphere?

It sounds like I should stop denigrating trivial bundles!
 

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fresh_42

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You confused me.
Yes. You can comb a hairy 3 sphere.
What about the hairy ball theorem, or as we call it, hedgehog theorem (D. Hilbert):

There is a tangential, continuous, nowhere vanishing vector field on ##\mathbb{S}^n##, if and only if ##n## is odd.
 

Geofleur

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If the tangent bundle of ##S^2## were globally trivial, then we could set it equal to ##S^2\times {\rm I\!R}^2## everywhere. But we could then set the values of the two component sets any which way we like. In particular, we could have a vector field ##(p,\textbf{v})## where ##v## is some constant non-zero vector and ##p## ranges over the whole sphere. But that would be a continuous, nowhere vanishing vector field on ##S^2##. Apparently, when ##n = 3## there's no problem with having such a vector field! Hence, what lavinia said is consistent with the hairy ball theorem as you quoted it :-⁠)
 

fresh_42

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If the tangent bundle of ##S^2## were globally trivial, then we could set it equal to ##S^2\times {\rm I\!R}^2## everywhere. But we could then set the values of the two component sets any which way we like. In particular, we could have a vector field ##(p,\textbf{v})## where ##v## is some constant non-zero vector and ##p## ranges over the whole sphere. But that would be a continuous, nowhere vanishing vector field on ##S^2##. Apparently, when ##n = 3## there's no problem with having such a vector field! Hence, what lavinia said is consistent with the hairy ball theorem as you quoted it :-⁠)
Sorry @lavinia , that is my regular mistake at this point. If I don't take care I confuse the index of the spheres with the dimension of the space they live in, not their own ... :oops:
And yes, I know the embedding isn't necessary, it is just the association seeing a ball, or a sphere.
 

lavinia

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The only spheres with trivial tangent bundles are the circle, ##S^3##, and ##S^7##. But every odd dimensional sphere can be combed.
A closed manifold that can be combed must have zero Euler characteristic. The Euler characteristic of the 2 sphere is 2 so it can not be combed. The torus and the Klein bottle can be combed and both have zero Euler characteristic.

In the case of the Klein bottle, the tangent bundle splits into a Whitney sum of a trivial line bundle and a non-trivial line bundle.

I am not totally sure about the converse. Every closed orientable manifold is combable if and only if its Euler characteristic is zero. This does not generalize to arbitrary sphere. bundles over manifolds. There are I think 3 sphere bundles over the 4 sphere which have no global section but still have Euler class equal to zero. I think the difference is that the structure group cannot be reduced to the orthogonal group.

The 3 sphere is also a Lie group so not only can it be combed but its tangent bundle is trivial.
 
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lavinia

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I thought a note of explanation of some of the terms on post #11 might be helpful.

If a vector bundle ##η## has a non-zero section ##s## then this section determines a 1 dimensional subbundle ##λ## whose fibers are all scalar multiples of ##s## at each point of the base manifold. Each of its fibers is a 1 dimensional linear subspace of the larger vector bundle ##η## and using a Riemannian metric one gets a complementary subbundle ##ω## so that the fibers of ##η## are the direct sum of the fibers of ##λ## and ##ω##. ##η## is said to be the Whitney sum of ##λ## and ##ω##.

A 1 dimensional vector bundle is called a line bundle. If a line bundle is determined by a global section then it is called trivial. But a line bundle might not have a global section. The simplest example is the Mobius band which is a non-trivial line bundle over the circle. One can prove that it has no non-zero section using the Intermediate Value Theorem.

A vector bundle is called trivial if it can be written as a Whitney sum of trivial line bundles. This is the case that @Geofleur described in the first post. This is different than a vector. bundle that is the Whitney sum of arbitrary line bundles since some of the line bundles in the sum may be non-trivial.

Two illustrative examples are the tangent bundles of the torus and the Klein bottle. For the torus, the bundle splits into a Whitney sum of two trivial line bundles and so is trivial but for the Klein Bottle one gets a Whitney sum of a trivial line bundle and a non-trivial line bundle.

A manifold is combable if its tangent bundle splits into a Whitney sum with a trivial line bundle. The entire bundle itself though may be non-trivial. So the torus and the Klein bottle are both combable - if combable is even a word - but the tangent bundle of the Klein bottle is not trivial. On the other hand the 2 sphere is not combable nor is any other orientable closed surface other than the torus.

- If a vector bundle has a Riemannian metric then one can consider the vectors of length 1. These form a sphere in each fiber and one gets what is called a sphere bundle. If the vector bundle has a non-zero section then dividing each vector in the section by its length produces a section consisting completely of vectors of length 1 and thus is a section of the corresponding unit sphere bundle. This way of looking at sections suggests a generalization which is to start out with a sphere bundle and look for sections of it. If the sphere bundle is oriented e.g. if it inherits an orientation from an oriented vector bundle, then there is an obstruction to the existence of a global section called the Euler class of the sphere bundle. This cohomology class must be zero in order for there to be a section. In the special case of the tangent unit sphere bundle of an oriented closed manifold, the Euler class is zero if and only if the Euler characteristic of the manifold is zero. This is a non trivial theorem and there are examples of non-tangent sphere bundles with zero Euler class that have no section.
 
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