# A About Chern number of U(1) principal bundle on a torus

1. Apr 28, 2017

### lichen1983312

The U(1) bundle on a torus is a important math setup for a lot of physics problems. Since I am awkward on this subject and many of the physics material doesn't give a good introduction. I like to put some of my understanding here and please help me to check whether they are right or wrong.
1. A one-form $A$ can be defined over the whole torus.
2. To define a connection one-form for this bundle, we need a Lie-algebra valued one-form on the torus. So I can simply define this form by adding an $i$ to $A$ as $iA$.
3. So the Lie-algebra valued local curvature two-form is $F = diA + iA \wedge iA = idA$
4. If there is no continuous section can be found. Both $iA$ and $F$ can only be well defined on local charts. In the overlapping parts ${U_i} \cap {U_j}$ the transition function is ${t_{ij}}(p) = \exp [i\Lambda (p)]$. So we have the transition $iA \to iA + id\Lambda (p)$ and $F \to t_{ij}^{ - 1}Ft_{ij}^{ - 1} = F$ and the Chern number is found as
$\int_T {\frac{i}{{2\pi }}trF = } \int_T {\frac{i}{{2\pi }}F} \in Z$

So $F$ is exactly $idA$ everywhere? is this right ?

If so, by directly appling stokes theorem
$\int_T {F = - i} \int_T {dA = } \int_{\partial T} {A = 0}$

Last edited: Apr 28, 2017
2. Apr 28, 2017

### lavinia

For a general $U(1)$ bundle there is not a global gauge field although the Chern class is always a global 2 form. The field strength is exact only locally so one can not use Stokes Theorem to conclude that its integral over the torus is zero. If the bundle is trivial, that is if it has a global section, then the gauge field is a global Lie algebra valued 1-form and the Chern class is zero.

A basic example of this type of phenomenon is the global 1 form $dθ$ on the circle. This form is not exact although locally it is the differential of the angle function $θ$. However, $θ$ cannot be defined continuously on the entire circle.

- The Chern class of the tangent bundle of the torus is zero since it is a trivial bundle. But there are infinitely many $U(1)$ bundles on the torus that are not trivial.

Last edited: Aug 3, 2017
3. Apr 28, 2017

### lichen1983312

Thanks very much.
You said "Chern class is zero" do you really mean Chern number ?

So what is the reason $\theta$ is not exact? Just because it cannot be globally defined?
Also, can you explain more about how two U(1) bundles on the same base space are different?

Last edited: Apr 28, 2017
4. Apr 28, 2017

### lavinia

Same thing, The Chern number is zero if and only if the Chern class is zero. Otherwise put, the Chern number is zero if and only if the Chern form is exact and this is true if and only if its cohomology class is zero in $H^2($Torus;$R)$. When one says that the Chern class is zero, one means that its cohomology class is zero.

Last edited: Apr 28, 2017
5. Apr 29, 2017

### lavinia

$dθ$ is not exact ( even though it is written as an exterior derivative. But this is just notation). One way to see that it is not exact is to observe that its integral over the circle is equal to $2π$. If it were exact its integral would be zero - by Stokes Theorem.

Last edited: Apr 29, 2017
6. Apr 29, 2017

### lichen1983312

Thanks, I am really confused about something here. Can I say that $dθ$ is not exact is because the circle cannot be parameterized completely by $θ$?

7. Apr 29, 2017

### lavinia

$U(1)$ bundles are characterized by their Chern class. Two bundles have the same Chern class(equivalently Chern number) if and only if they are bundle isomorphic.

For example, on the 2 sphere, two non-isomorphic $U(1)$ bundles are the tangent bundle and the Hopf bundle. The tangent bundle has Chern number 2 and the Hopf bundle has Chern number 1.

Split the 2 sphere into two hemispheres along the equator. Since each hemisphere is a topological disk and since $U(1)$ is a topological circle, the bundle restricted to each hemisphere is homeomorphic to the Cartesian product $D^2×S^1$ which is a solid torus so the $U(1)$ bundle splits into two solid tori. These are pasted together along their boundaries which are just two regular tori (not solid). These boundary tori lie above the equator of the sphere. So a $U(1)$ bundle over the 2 sphere is obtained from two solid tori that are pasted together along their boundaries. Different ways of pasting these boundaries together produce different $U(1)$ bundles.

It is a good exercise to examine these pasting maps for the cases of the tangent bundle and the Hopf bundle.

Last edited: Apr 29, 2017
8. Apr 29, 2017

### lavinia

Yes. $dθ$ is the 1 form that assigns a value of 1 to the positively oriented unit tangent vectors of the circle at each point. So integrating over an arc gives the arc length. Arc length does not define a continuous function on the entire circle since if the arc goes once around the circle the integral is $2π$ but if the arc goes nowhere, the integral is zero.

Last edited: Apr 29, 2017
9. Apr 29, 2017

### lichen1983312

So, can I say that I can always have a globally defined one-form $A$ for an arbitrary manifold, as long as in the overlapping area of different charts it transform according to the transition function? Its exterior derivative $dA$ should be also globally defined? right ?

10. Apr 29, 2017

### lichen1983312

Thanks very much.

11. Apr 29, 2017

### lavinia

One can always have a globally defined form of any degree just by defining a form on a coordinate neighborhood and extending it to the whole manifold by multiplying it by a smooth function that equals zero outside of the neighborhood.

12. Apr 29, 2017

### lichen1983312

Yes, that is generally possible right?

13. Apr 29, 2017

### lavinia

Yes. The 1 form that is identically zero everywhere is global as well

Here is an example. Choose a non-negative smooth function on the circle that is zero outside of a small interval around 1 and which is strictly positive on a subinterval on non-zero length. The graph of this function looks like a little bump surrounding 1 that flattens out completely outside of the interval. Multiply $dθ$ by this function. This produces a new 1 form that is identically zero outside of the interval. However, the integral of this 1 form over the circle is not zero since the function is positive on an interval of non-zero length. Therefore by Stokes Theorem this form is not exact.

Last edited: Apr 29, 2017
14. Apr 29, 2017

### lichen1983312

Sorry I was trying to say a every where non-zero form. So if I go back to my original post, and let $A$ be such a globally defined one-form on the torus. Since the bundle is not necessarily trivial. On certain charts the gauge potential has the form $iA + id{\Lambda _1} + id{\Lambda _2} \cdots$, but the field strength is globally defined as $d(iA + id{\Lambda _1} + id{\Lambda _2} \cdots ) = idA$.
is this right ?

15. Apr 29, 2017

### lavinia

This is how I understand $U(1)$ gauge potentials. One has a connection 1 form $ω$ defined globally on the tangent space to the $U(1)$ bundle itself - not on the tangent space to the manifold. This is a Lie algebra valued 1 form. Given a local section of the bundle one can pull the connection 1 form back to the tangent bundle of the manifold to get what is called a gauge field. This field in general is only defined locally since there may not be a global section of the bundle. Given a different local section one gets a different gauge field over a possibly different local neighborhood.

The exterior derivative of the connection 1 form on the $U(1)$ bundle is the curvature 2 form of the connection. Pulling this form back by the local section gives the field strength of the gauge potential. One can show - and I think you did - that the different field strengths over different local neighborhoods piece together to give a global 2 form. This form represents the Chern class of the $U(1)$ bundle and its integral over the torus - or for that matter any complete orientable surface without boundary e.g. the sphere - is the Chern number of the bundle.

Notice though that the gauge fields themselves may not piece together to give a global 1 form and this is where I think your question comes from. You said in the original post

"So $F$ is exactly $idA$ everywhere? is this right ?"

In general this is wrong because the gauge potential $A$ may not be defined globally but only locally. If however, $A$ is in fact defined globally then you are right to conclude that the Chern number is zero.

- BTW; Notice that this is the same kind of situation as with $dθ$. Locally one has an angle function $θ$ whose exterior derivative equals $dθ$ but there is no global function on the circle whose exterior derivative equals $dθ$. In the same way one has a global 2 form $dA$ defined on the torus which is locally the exterior derivative of a 1 form. But globally it is not an exterior derivative unless the $U(1)$ bundle is trivial.

Last edited: Apr 30, 2017
16. Apr 30, 2017

### lichen1983312

Thanks for your explanation, that is very helpful. I think the mistake in my example is that I tried to build a globally defined gauge potential (one-form) on the base space first, then I used it to define the connection one-form on the bundle. This is not valid at all. One should always start from a connection-one form on the bundle first and then push it back to the base manifold.
Is this right?

17. Apr 30, 2017

### lavinia

Yes. You got it.

BTW: This helped me to understand the Chern class of a $U(1)$ bundle. $U(1)$ topologically is a circle so it is a 1 dimensional sphere - as a manifold. A $U(1)$ bundle is an example of a sphere bundle, a fiber bundle whose fiber is homeomorphic to a sphere - in this case a 1 sphere. The Chern class of a $U(1)$ bundle is a special case of a characteristic class that is defined for oriented sphere bundles where the fiber sphere can be of any dimension. This characteristic class is called the Euler class of the bundle.

For smooth bundles over smooth manifolds, the Euler class is characterized by the following construction: First one can show that for an oriented k-sphere bundle there is a k-form $ω$ defined on the total space of the bundle whose integral over each fiber sphere is equal to 1. For $U(1)$ bundles $ω$ can be chosen to be the connection 1 form appropriately normalized. The exterior derivative of $ω$ for any k-sphere bundle is a smooth $k+1$ form and remarkably it is always the pull back under the bundle projection map of a globally defined $k+1$ form on the base manifold. That is: $dω = π^{*}C$ and $C$ is called the Euler form of the sphere bundle. For a $U(1)$ bundle $dω$ is the curvature 2 form of the connection and $C$ is the Chern form.

Notice that $C$ may not be exact even though its pull back to the total space is equal to $dω$. But for any local section $s$ of the bundle $C = ds^{*}ω$ so $C$ is locally exact but in general not globally. We saw this with the Chern form of the $U(1)$ bundle.

Now suppose that there actually is a global section of the sphere bundle. Then $C = s^{*}dω = ds^{*}ω$ so $C$ is exact. So the Euler class is an obstruction to a section of the sphere bundle. That is: if there is a section then the Euler class must be homologous to zero i.e. any differential form that represents it must be exact.

In general, one can not integrate the Euler class over the base manifold since $C$ is a $k+1$ form and the dimension of manifold may not be equal to $k+1$, for instance a $U(1)$ bundle over the 5 sphere. However if the manifold's dimension equals $k+1$ then it can be integrated over the manifold to yield the Euler number of the bundle. For the case of a $U(1)$ bundle over a 2 dimensional surface the Euler number is the same as the Chern number.

For the case of the tangent bundle of a smooth manifold, the Euler number is a topological invariant called the Euler characteristic.

Notice that if there are two $k$ forms $ω_1$ and $ω_2$ then their difference defines a global $k$ form on the base manifold. It follows that the difference of the corresponding Euler forms $C_1$ and $C_2$ is exact so the Euler class is independent of the choice of $ω$. For a $U(1)$ bundle this means that the Chern class is independent of the connection. So it is an invariant of the bundle and is independent of the bundle's geometry.

Last edited: May 7, 2017
18. Apr 30, 2017

### WWGD

Lavinia, can you recommend a good source for this material, (diff. topology, I guess)? Maybe Bott and Tu?

19. Apr 30, 2017

### lavinia

Yes Bott and Tu for Euler classes using differential forms. For Characteristic classes derived from connections on vector bundles - Milnor's Characteristic Classes Appendix C - For gauge fields try this link

https://empg.maths.ed.ac.uk/Activities/GT/Lect1.pdf

Bott and Tu do the Euler class for smooth oriented sphere bundles using de Rham theory and Czech cohomology. Really good. They show examples of sphere bundles whose Euler class is zero but have no section.

For the Euler class using differential geometry on a Riemannian manifold Chern's original paper. I think it is called "A Simple Intrinsic Proof of the Gauss-Bonnet Theorem." Also Milnor's Appendix C. If the metric is not positive definite I am not sure. The Pfaffian form probably has some analogue for non-positive definite metrics.

Last edited: Apr 30, 2017
20. May 1, 2017

### lichen1983312

Can you help me to check this example? It is one of the reasons I post this thread. It is the magnetic monopole problem.
If there is a magnetic monopole, it should produce a magnetic field $\vec B = g\vec r/{r^3}$ so that $\int_S {\vec B \cdot d\vec S} = 4\pi g$, where $4\pi g$ is a constant. It is also known that in physics there should be a vector potential ${\vec A}$ such that $\vec B = \nabla \times \vec A$. As you can expect, there is no globally defined ${\vec A}$ can be found. So people use two vector potentials ${{\vec A}_N}$and ${{\vec A}_S}$.
${{\vec A}_N}$ is defined everywhere except the south pole with components
$A_N^x = \frac{{ - gy}}{{r(r + z)}}\,$, $A_N^y = \frac{{gx}}{{r(r + z)}}$, and $A_N^z = 0$.
${{\vec A}_S}$ is defined everywhere except the north pole with components
$A_S^x = \frac{{gy}}{{r(r - z)}}\,\,$, $A_S^y = \frac{{ - gx}}{{r(r - z)}}$ and $A_S^z = 0$
So if think $A$ is a one-form
${A_N} = A_N^xdx + A_N^ydy$ and ${A_S} = A_S^xdx + A_S^ydy$
by switching to spherical coordinates
${A_N} = g(1 - \cos \theta )d\varphi$
${A_S} = - g(1 + \cos \theta )d\varphi$
and the curvature is globally defined
$d{A_N} = d{A_S} = g\sin \theta d\theta \wedge d\varphi$
so the books says if we put an Lie algebra factor "i" in front of "A" , "iA" becomes a Lie algebra valued one-form. "idA" becomes the field strength.
$i{A_N} - i{A_S} = id(2g\varphi )$
is also a valid transition property.
So the books says we have are dealing with a $U(1)$ bundle and in this case it is non-trivial.

So my question is
Is this a case that we construct a U(1) bundle from properties from the base manifold?
Can I just conclude that if I have
1, locally defined Lie algebra-valued one-form ${A_i}$ on the base manifold.
2, ${A_i}$ have a transitions property ${A_i} = {A_j} + id\phi$.
3, $dA$ is globally defined
there must be a well defined U(1) bundle corresponding to these properties?