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Why chern number must be an integer?

  1. Jun 17, 2010 #1
    i am a student of physics

    i do not know much about the chern number

    i wonder why the chern number must be an integer

    any good reference?
  2. jcsd
  3. Jun 17, 2010 #2
    I think this may answeryour question; maybe someone else can critique it.

    Let's see; let's assume we are using integers as the coefficient group G.
    The n-th chern class of (the tangent bundle of) a
    m-manifold, is, by def. an element of H<sup>2n</sup>(M;G) , i.e., the 2n-th cohomology
    group of M. So , strictly speaking, the Chern class is a cochain representative of
    some cohomology class, so an element of Hom( H_2(M;G),G) , by (vector space)
    duality of homology with resp. to cohomology. So, if G =integers, you get a map into
    the integers as your Chern class, and so the value of the Chern class must be an

    Does that answer your question.?. I know physicist often complain about mathematicians
    using jargon that obscures instead of clarifying. Let me know if this was the case here,
    and I will try to clarify.
  4. Jun 17, 2010 #3
    thanks a lot

    but i hope it can be as concrete as possible

    take a 2 dimensional closed surface in three dimensions, ok?
  5. Jun 18, 2010 #4
    Intuitively, the Chern number can be thought of how much a complex fiber 'winds' around a manifold. This is an imperfect example, but it builds the intuition well. As a first example, consider the Mobius strip. We can think of it as having the base space S^1 with the line segment [0,1] as a fiber that 'winds around' the base space. Obviously, in order for the fiber to 'meet up' with itself where it completes a circuit around the base space, it will have to 'wind around' the base space an integer number of times. Otherwise you will end up with a sort of angular mismatch between the pieces if you wind around a non-integer number of times.

    Now, since Chern numbers only talk about complex fibers, this isn't a perfect example. But, the idea is the same. If your fiber doesn't wind around the base space an integer number of times, you'll end up with a mismatch when you try to glue it all together in a cohesive whole. Does that help? I'm a physics student too, and this is how I've always thought about it.
    Last edited: Jun 18, 2010
  6. Jun 18, 2010 #5
    thanks a lot

    is there any good reference for physics student?

    i am curious about why the chern number associated with the berry curvature must be an integer
  7. Jun 19, 2010 #6
    I learned about Chern numbers from Topology, Geometry, and Physics by M. Nakahara. It's very well written, and aimed at a physics audience. I'm an undergraduate physics major and I found it very accessible.

    The reason that the Chern number linked to Berry's phase must be an integer becomes clear when you consider the phase of a wave function. Berry's phase, of course, arises from curvature in the parameter space of possible Hamiltonians. Consider adiabatic transport of an eigenstate around a small loop in the parameter space. Assuming some reasonable 'niceness' conditions on the parameter space, when you complete the loop, the particle will be in the same eigenstate it started as, with a change of phase [tex]e^{i\phi_1}[/tex]. By Stoke's theorem, this [tex]\phi_1[/tex] should be some function of the area inside of the loop taken. Then, by adiabatically transporting the eigenstate in the opposite direction around the loop, the change in phase should be [tex]e^{-i\phi_2}[/tex], where [tex]\phi_2[/tex] is some function of the area of the parameter space outside of the loop. But, going around the loop clockwise should result in the same change of phase as going around the loop counterclockwise. So we set

    [tex]e^{i\phi_1} = e^{-i\phi_2}[/tex]

    Setting [tex]\phi = \phi_1 + \phi_2[/tex], we get the condition [tex]e^{i\phi} = 1[/tex], meaning that [tex]\phi[/tex] MUST be an integer multiple of [tex]2\pi[/tex]. This integer is the Chern number.

    What is really going on here is that you have a fiber bundle with a base space consisting of the parameter space, and the fiber at each point is the associated eigenspace with the Hamiltonian at that point (perhaps with a cutoff at some energy to avoid the complication of an infinite dimensional fiber). Then, imagine the evolution of the eigenvector as it takes a trip around the parameter space. Assuming some 'niceness' conditions on the parameter space (namely that there is no degeneracy, energy bands never touch, and the dimension of the eigenspace never changes), then you can imagine the eigenvector traveling along with the frame of the eigenspace as it is transported around the manifold. In the presence of say, a magnetic field, the fiber twists around the parameter space. But, in order for it to be able to be 'glued' back to itself appropriately, it has to twist around the parameter space an integer number of times. Otherwise, when you imagine the eigenvector traveling along the frame of the eigenspace, it will end up in the wrong spot when you travel along a loop. I.e., you'd start with some state [tex]\psi[/tex] and instead of transporting [tex]\psi[/tex] around a loop and ending up with [tex]e^{i\phi} \psi[/tex] you might get some other eigenstate [tex]\xi[/tex]. This makes no physical sense, so it must be that the fiber 'twists' around the parameter space an integer number of times.
  8. Jun 19, 2010 #7

    thanks a lot

    i now understand

    i am shamed that i am phd student in the fifth year
  9. Jun 19, 2010 #8
    There's no shame in asking questions! The only reason that I know this well is because I ask lots and lots of questions :) It's not like anyone is born knowing this junk.
  10. Jun 20, 2010 #9
    Maybe a more egghead explanation than Monocles' nice physical one -- I include
    it here because I think it simplifies my first reply -- is that
    a Chern class is an element of the linear dual of Homology, i.e., it is a linear
    integer-valued ( if we work over the integers) map from a Homology class to the
    integers. So your Chern class is --by this def. of a Chern class -- an integer-valued
    linear map.

    Still, it would be nice to see what happens if you work with different coefficients;
    of course, you could then use the universal coeff. theorem, but I don't think you would
    then necessarily get an integer. Anyone else know.?

    It would be nice to understand how/why the physical and mathematical approaches
    agree with each other.
  11. Jun 22, 2010 #10


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    Chern classes are integer cohomology classes. On an oriented manifold the numbers must be integers.

    The remarkable fact is that Chern classes can be expressed as differential forms derived from the curvature 2 form. These are real cohomology classes but the numbers they produce are always integers.
  12. Jul 5, 2011 #11


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    I don't like when mathematics try to theorem physics, that some effect is forbidden.
    There exists forbidden graphene.
    There exists unexpected fractional Hall effect.
    There exists unexpected fractional charge in quarks.

    I don't see any physical proof, that chern number can't be fractional.
    If anybody claim the proof, you are welcome to point out.

    We can have exact theorems in mathematics, but i don't think that mathematics has the right to rule physics. Why should the physical world follow mathematical conditions?
    Last edited: Jul 5, 2011
  13. Jul 7, 2011 #12


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    we do not know why the observed world obeys mathematical laws but it does. And mathematical theorems can be used to predict new observations.
  14. Jul 7, 2011 #13
    This doesn't make any sense, since a Chern number is a precisely defined mathematical quantity and that does not require a "physical proof" (whatever that means). And math requires that the Chern number must be an integer (because it comes from a integer cohomology).

    What you probably meant was, you don't see a proof of why the plateaus in Quantum Hall (QH) systems can't be fractional. These plateaus are physical (in contrast to a concept like Chern number) and therefore this would require a "physical proof".

    In this case, the logic is that under certain assumptions (such as non-interacting electrons) one can derive that the QH plateaus (physical) are given by the Chern number of a certain fiber bundle (mathematical). This then says that the plateaus MUST be quantized as integers, AS LONG as the assumptions are fulfilled. If fractional plateaus are observed, then it means that one or more assumptions are not fulfilled (such as interaction between electrons must be taken into account). It DOES NOT mean that it is a "physical proof" that the Chern number can be fractional (cause it can't!), it means that the plateaus are NOT directly given by Chern numbers when interactions are included.

    A mathematical description of a physical phenomena relies on many assumptions, and is ONLY valid as long as the assumptions are. Therefore the fact that Chern numbers are integers (which is a mathematical fact) does NOT mean, fractional Quantum Hall effect cannot exist.

    This is more a philosophical question (see http://en.wikipedia.org/wiki/The_Unreasonable_Effectiveness_of_Mathematics_in_the_Natural_Sciences" [Broken]), all we know is that mathematics describe the physical world (as long as we understand the limitations of the mathematical description) and abstract mathematical theorems have provided numerous highly non-trivial predictions.
    Last edited by a moderator: May 5, 2017
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