# Integration along a loop in the base space of U(1) bundles

Let ##P## be a ##U(1)## principal bundle over base space ##M##.
In physics there are phenomenons related to a loop integration in ##M##, such as the Berry's phase
##\gamma = \oint_C A ##
where ##C(t)## is a loop in ##M##, and ##A## is the gauge potential (pull back of connection one-form of ##P## on ##M##).
It seems that if we can find an ##A## that is smooth and single valued on the loop, ##\gamma ## is well defined. Physics literatures always assume this is true. However, we know in general there may not be a smooth and single valued ##A## found over ##M##. Then my question is what determines there is a good ##A## defined on the loop?

My guess is that since ##f:C \to M## is a smooth map, so the topology of the pullback bundle ##{f^ * }P## on ##C## determines whether ##\gamma ## can be well defined on ##M##? i.e. if ##{f^ * }P## is trivial, then a smooth and single valued ##A## can be defined on ##C## in ##M##.

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lavinia
Gold Member
A couple thoughts.

- The induced bundle over the circle will be a smooth surface that is a circle bundle over a circle. There are only two such surfaces, the torus and the Klein bottle. Which is it?

- Over the circle minus a point,pp, the bundle is trivial. Near the two ends, a section approaches two possibly different points in the fiber over the omitted point,pp. Adjust the section by the action of U(1)U1 on the fibers so that the two points match up at pp and which continuously rotates to become the identity outside of a small interval around pp.

- Every principal bundle(over a paracompact space) has a connection.(standard theorem)

- Every principal ##U(1)## bundle is uniquely characterized by its Chern class. If the Chern class is zero, the bundle is trivial. The Chern class is an integer cohomology class in the second cohomology group of the base space.

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A couple thoughts.

- The induced bundle over the circle will be a smooth surface that is a circle bundle over a circle. There are only two such surfaces, the torus and the Klein bottle. Which is it?

- Over the circle minus a point,##p##, the bundle is trivial. Near the two ends, a section approaches two possibly different points in the fiber over the omitted point,##p##. Adjust the section by the action of ##U(1)## on the fibers so that the two points match up at ##p## and which continuously rotates to become the identity outside of a small interval around ##p##.

- Every principal bundle(over a paracompact space) has a connection.(standard theorem).

- Every principal ##U(1)## bundle is uniquely characterized by its Chern class. If the Chern class is zero, the bundle is trivial. The Chern class is an integer cohomology class in the second cohomology group of the base space.

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Hi Lavia:
Thanks very much, but what character characterize the circle bundle on a circle, I only know little about the Chern class, which seems to be only defined on a even denominational base space ?

lavinia
Gold Member
Hi Lavia:
Thanks very much, but what character characterize the circle bundle on a circle, I only know little about the Chern class, which seems to be only defined on a even denominational base space ?
- The Chern class is defined for any complex line bundle/principal ##U(1)## bundle.

- One knows from the classification of surfaces that the only two surfaces that fiber over the circle are the torus and the Klein bottle. One proof is to use the theorem that the Euler characteristic of a fiber bundle over a compact path connected space with compact fiber is the product of the Euler characteristics of the fiber and the base space. The circle has Euler characteristic zero so the bundle has Euler characteristic zero. The only two surfaces with zero Euler characteristic are the torus and the Klein bottle.

One must prove that if the circle bundle is a ##U(1)## bundle then it must be a torus.

- A torus is the Cartesian product of two circles ##S^1×S^1## and so is a trivial circle bundle over the circle. ##U(1)## acts on the fibers by complex multiplication.

The Klein bottle is not a product of circles. It is the quotient space of the torus by the action of the group ##Z_{2}## that rotates the first circle by 180 degrees and reflects the second circle around an axis. This action is orientation reversing and the Klein bottle is not orientable. As a circle bundle over the circle, one can show using the Intermediate Value Theorem that the bundle does not have a section.

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Hi Lavinia, Thanks very much, that is very helpful!!

lavinia
Gold Member
Hi Lavinia, Thanks very much, that is very helpful!!
You are welcome.

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The Chern class is defined for any complex line bundle/principal U(1)U(1)U(1) bundle.
I have to ask this again, should the Chern class be only defined on a bundle whose base space is even dimensional?
And are the Chern classes defined on a ##U(1)## bundle on a circle?

lavinia
Gold Member
I have to ask this again, should the Chern class be only defined on a bundle whose base space is even dimensional?
And are the Chern classes defined on a ##U(1)## bundle on a circle?
Explain why you are asking again. What is the confusion?

Explain why you are asking again. What is the confusion?
I mean should the Chern class be a even dimensional form like 2-form, 4-form..., and the circle is one dimensional.
or you mean the 0-form is well defined?
May be I am confused about the definition of Chern number. Does the U(1) bundle on a cirlce have a Chern number?

lavinia
Gold Member
I mean should the Chern class be a even dimensional form like 2-form, 4-form..., and the circle is one dimensional.
or you mean the 0-form is well defined?
May be I am confused about the definition of Chern number. Does the U(1) bundle on a cirlce have a Chern number?
I suspected this may have been the question.

The Chern class is an integral 2 dimensional cohomology class. It exists for any complex line bundle. It may be represented by a closed 2 form if the bundle is smooth.

The Chern number is the integral of the Chern class over a closed two dimensional manifold.

In general, a manifold may have many closed surfaces embedded inside it. One can integrate the Chern form over any one of them. But these are not called Chern numbers.

Since every 2 form on the circle is zero, the Chern class of any ##U(1)## bundle over the circle is trivial. In homology terms, ##H^2(circle;Z)=0##.

I suspected this may have been the question.

The Chern class is an integral 2 dimensional cohomology class. It exists for any complex line bundle. It may be represented by a closed 2 form if the bundle is smooth.

The Chern number is the integral of the Chern class over a closed two dimensional manifold.

In general, a manifold may have many closed surfaces embedded inside it. One can integrate the Chern form over any one of them. But these are not called Chern numbers.

Since every 2 form on the circle is zero, the Chern class of any ##U(1)## bundle over the circle is trivial. In homology terms, ##H^2(circle;Z)=0##.
Thanks very much for the explaination.

lavinia
Gold Member
Thanks very much for the explaination.
You are welcome.

BTW: If one has a complex vector bundle of higher complex dimension ##n## then there is a Chern cohomology class in each even dimension up to ##2n##. So a complex 2 plane bundle will have a Chern class in ##H^2(M;Z)## and ##H^4(M;Z)##. A 3 plane bundle will also have a 6 dimensional class.

A complex 2 plane bundle over a closed 4 manifold will have two Chern numbers. One is the integral of the second Chern class (the one in ##H^4(M;Z)##) and the other is the integral of the first Chern class (the one in ##H^2(M;Z)##) wedge product with itself. This idea generalized to all dimensions. If the bundle is the tangent bundle of the manifold then these are called Chern numbers of the manifold.

The tangent bundle of any closed oriented surface( e.g. a torus or a sphere )is a complex line bundle. The Chern class of the tangent bundle integrates to the Euler characteristic of the surface.

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