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A Why the Chern numbers (integral of Chern class) are integers?

  1. Oct 29, 2016 #1
    I am a physics student trying to self-learn Chern numbers and Chern class. The book I am learning (Nakahara) introduces the total Chern class as an invariant polynomial of local curvature form ##F##
    ## P(F) = \det (I + t\frac{{iF}}{{2\pi }}) = \sum\limits_{r = 0}^k {{t^r}{P_r}(F)} ##
    and each ##{P_j}(F)## defines the j-th Chern class##{c_j}(F) \in {H^{2j}}(M)##

    The book didn't mention anything about the Chern number. According to some other material I found (may be wrong), the Chern number is defined as an integral over 2##r##-cycle,

    ##\int_\sigma {{c_{{j_1}}}(F)} \wedge {c_{{j_2}}}(F) \cdots {c_{{j_l}}}(F) ##

    where ##{j_1} + {j_2} + \cdots {j_l} = r##

    The material also said that this integral is always an integer, which is used in many context of physic. Due to my limited knowlege, I cannot see how this is proved and I cannot find some reference that is easy enough to me. So can anybody help ?
     
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  3. Oct 29, 2016 #2

    lavinia

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    Chern classes are defined for any complex vector bundle as cohomology classes with integer coefficients. Their value on any cycle is therefore an integer.

    The proof that the invariant polynomials applied to the curvature 2 form of a connection are the Chern classes of a complex vector bundle over a manifold requires a series of steps -at least for the proof that I have seen.

    These are:

    1) Show that the invariant polynomials on local charts fit together across charts to produce a global differential form.
    2) Show that this differential form is closed.

    Steps 1 and 2 show that the invariant polynomial defines a de Rham cohomology class. This is a cohomology class with complex coefficients.

    3) Show that the cohomology class of this closed form is independent of the connection. The proof of this is straightforward using a homotopy between the two connections and the theorem that homotopic cocycles are cohomologous.

    This means that the cohomology class depends only on the vector bundle (up to isomorphism).

    4) Show that the cohomology class of this differential form is natural. That means that the cohomology class of the invariant polynomial on the induced bundle is the pull back of the cohomology class of the invariant polynomial on the target bundle. In fact the invariant polynomial on the induced bundle with the induced connection is the pullback of the invariant polynomial on the target bundle.

    This shows that the polynomial defines a characteristic class: That is: it is natural and is determined by the isomorphism class of the complex vector bundle alone. A general theorem then says that all characteristic classes of complex vector bundles are polynomials in the Chern classes.

    5) Once one knows what all of the invariant polynomials are one can show that they are each a constant multiple of some Chern class. By naturality this constant is the same for all bundles and thus can be determined from a single bundle. For instance the invariant polynomial for a complex line bundle ##γ## just the ##i## times curvature 2 form ##Ω_{12} ## and the only Chern class of a complex line bundle is the first Chern class. Thus ##iΩ_{12} = αc_{1}(γ)##. One can calculate ##α## on a single example of a complex line bundle for instance the tangent bundle of the 2 sphere (with any connection). By the Gauss-Bonnet theorem one knows that the integral of the curvature 2 form is ##2π## time the Euler characteristic of the sphere. One also knows that the first Chern class for any complex line bundle is its Euler class and that for the tangent bundle of any closed orientable surface, its value on the fundamental cycle is the Euler characteristic. (The tangent bundle of any closed orientable surface is a complex line bundle). So ##α = 2πi##.

    For complex vector bundles of higher dimension, the proof reduces to the case of a Whitney sum of complex line bundles and the product formula for Chern classes.

    I know that this answer doesn't really give the proof. But that would take a lot of space.

    BTW: For real vector bundles this proof immediately shows that the Pontryagin classes are represented by invariant polynomials, this because Pontryagin classes are the Chern classes of the complexified real bundle.

    For the Euler class of a real vector bundle one needs a different invariant polynomial. It is called the Pfaffian. In general the Euler class is not a Chern class and is defined for any real vector bundle.
     
    Last edited: Oct 30, 2016
  4. Oct 30, 2016 #3

    lavinia

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    Chern numbers are the values of products of Chern classes on the fundamental cycle of the manifold. A complex vector bundle over a manifold will have more than one Chern number unless it is a line bundle.
     
  5. Oct 31, 2016 #4
    Thanks for such a detailed explanation. I am trying to digest it now. But after looking at your post and some other mateirals I feel that a full understanding of the proof requires a much broader background knowledge than I have. I do have a lot of questions about your post. But I like to ask this question first. Since I never took a class of algebraic topology, all I know about cohomology is some basics of de Pham cohomology, do you think my backgroud is enough to understand it?
     
  6. Oct 31, 2016 #5

    lavinia

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    You can avoid the algebraic topology but you do need to know some geometry and some theory of vector bundles. This will at least get to showing that the invariant polynomials give characteristic classes that integrate to integers. For smooth complex vector bundles you could take that as the definition of Chern classes.

    I recommend that as a physicist you do learn the algebraic topology because cohomology with real or complex coefficients (de Rham theory) does not cover all of the physically interesting cases. For instance, Chern-Simons invariants are cohomology classes with ##R/Z## coefficients. Also Chern classes are invariants of complex vector bundles not of manifolds and are defined for any topological space.

    I am happy to guide you through everything.

    - Here is something to think about. For two different connections on a complex vector bundle, an invariant polynomial applied to the curvature two form matrix gives two different differential forms. If this is true, then what does it means to say that these two different forms represent the same Chern class?
     
    Last edited: Nov 2, 2016
  7. Nov 17, 2016 #6

    mathwonk

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    The simplest case is a complex line bundle over a compact Riemann surface. Any non trivial holomorphic section of such a bundle has isolated zeroes. In this case the chern number is just the number of zeroes (with multiplicities), hence an integer.

    Note that since the quotient of two such sections is a global meromorphic function, the fact that such a function ahs the same number of zeroes and poles, proves that all holomorphic sections have the same number of zeroes.

    This phenomenon is a reflection of the comment of Lavinia relating chern class to euler class.

    Another nice way to look at the chern classes is that they are always pulled back from a grassmannian. I.e. a grasmannian has a natural vector bundle on it, and every vector bumndle arises as a pull abck from these naural bundls via a classifying map from your manifold to the grassmannian.

    Then by the naturality of chern classes, the chern classes of your manifold must also arise as pull backs. Thus since the chern classes on a grassmannian are integral, so are all chern classes.
     
  8. Nov 17, 2016 #7
    Thanks very much !
     
  9. Nov 17, 2016 #8

    mathwonk

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    So, and perhaps this is Lavinia's point of view, the chern classes are integral by definition, or by their universal construction. So the question becomes why does that prescription using differential geometry and curvature, actually compute them. I.e. any deRham class that computes chern classes must be integral because the chern classes are integral. So one tries to give a way to recognize the chern classes by their universal properties. Then it turns out it is sufficient to show your deRham class gives the chern classes for some restricted set of manifolds, such as grassmannians, and then one just computes them there to show this. Perhaps by the "splitting" principle one can even reduce to the case of line bundles.
     
  10. Nov 17, 2016 #9
    can you be more specific about "the chern classes are integral by definition"?
     
  11. Nov 17, 2016 #10

    mathwonk

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    well of course it depends on what definition you choose. they are not integral in any obvious way by your definition, so my point is that yours is not the "correct" definition, or not the most natural one. i just meant there are many definitions and some of them include integrality, so thn the task is to equate other definitions with these.

    e.g. if O is the sheaf of holomorphic functions on a manifold, and O* is the subsheaf of units, i.e. never vanishing holomorphic functions, then the group of holomorphic line bundles on a manifold M, up to isomorphism, is the cohomology group H^1(M,O*). Then the exponential map defines a sheaf map from holomorphic functions to never vanishing ones hence a map of sheaves from O-->O*.

    If we multiply by 2πi, the kernel of this map is integer valued functions so we have an exact sequence of sheaves 0-->Z-->O-->O*-->0. Taking cohomology yields a long exact sequence of cohomology groups, including the map H^1(M,O*)-->H^2(M,Z), which is by one definition, the chern clas map. thus by this definition the chern class of a line bundle in H^1(M,O*) is an element of the integral cohomology group H^2(M,Z).


    Another point of view is that we choose any holomorphic section of our line bundle, (assuming they exist), and this defines a subvariety Z of codimension 2 (complex codimension one) in our manifold. then by another definition the chern class is the operation of intersecting with this subvariety, acting on the integral homology group of 2-cycles. Again this definition counts intersection numbers, and gives an element of integral cohomology, hence is integral by definition.

    Another definition is to define the chern classes only of the grassmannian, and one chooses them to be certain natural integral cohomology classes. then any vector bundle on M defines a map of M into the grassmannian, unique up to homotopy, and to deine the chern classes on M one just pulls back those universal ones. Since the pull back of an integral class is integral, again this definition gives integral classes.

    So then one asks how to compute these classes via deRham cohomology, and one comes up with the formulas you are being shown, masquerading as a definition. The problem then is to prove those formulas do compute the natural chern classes.

    the way to do this is to prove that the natural definition satisfies certain natural propwrties that make them unique. then anything having thiose properties will equal the natural ones. so if we prove that your deRham definition does satisfy the natural pullback property, we only need to check that yours are integral on a grassmannian.


    as a simple example, how does one check that the differential form
    (1/2π)(-ydx/(x^2+y^2) + xdy/(x^2+y^2)) is integral on the plane? the proof is to show that if theta = arctan(y/x) is the (multivalued) angle function, then this (single valued) form is dtheta. And it counts 1/(2π) times the change in angle of a closed curve around the origin, i.e. it counts the winding number , which is an integer.

    upshot: to show a given deRham class is integral, equate it with a geometrically natural integral class.

    I do not know how chern actually arrived at his definition, but i am guessing he took a geometric definition and computed it in deRham cohomology.
     
    Last edited: Nov 17, 2016
  12. Nov 17, 2016 #11

    lavinia

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    One can define Chern classes inductively as pull backs of Euler classes of a sequence of complex vector bundles derived from the original bundle and since Euler classes are integer cohomology classes so are Chern classes. The top Chern class is just the Euler class. of the original bundle itself.

    Alternatively one can define Chern classes and their products as the pull backs under the bundle classifying map of the integer cohomology classes of the Grassmann manifold of complex n-planes. The integer cohomology of the complex Grassmann is a polynomial algebra with one generator in each even dimension. Each of these generators pulls back to a Chern class under a bundle classifying map. The odd dimensional integer cohomology of the complex Grassmann manifold is zero. This rules out the possibility of odd dimensional Chern classes and odd dimensional products of Chern classes.

    Technically the invariant polynomials in the curvature 2 form matrix are de Rham cohomology classes. They are the image of the Chern classes under the mapping ##H^{*}(;Z)→H^{*}(;R)## induced by the coefficient inclusion ##Z→R##
     
    Last edited: Nov 23, 2016
  13. Nov 24, 2016 #12

    lavinia

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    A note on Euler classes.

    This will show how Euler classes may be defined as integer cohomology classes.

    Euler classes are not defined for all vector bundles but only for oriented vector bundles. An orientation can be thought of as a consistent choice of orientations of each of the fibers. This in turn can be thought of as a consistent choice of a generator of each integer cohomology group ##H^{k}(V_{p},V_{p}- 0;Z)## for the fiber ##V_{p}## above each point ##p## in ##M##. ( Note that ##H^{k}(V_{p},V_{p}- 0;Z) ≈ Z## so it has two generators.)

    Given an orientation a fundamental theorem proves that there is a unique cohomology class in ##H^{K}(V,V-0;Z)## that pulls back to the chosen generator of ##H^{k}(V_{p},V_{p}-0;Z)## under the fiber inclusion map ##V_{p} →V##. (Here ##V-0## is the total space of the vector bundle minus all of the zero vectors.)

    The Euler class is defined as the pull back to the base space ##M## of this uniquely determined cohomology class in ##H^{k}(V, V-0);Z)## . The pull back involves two maps:

    ##H^{k}(V, V-0;Z) →H^{k}(V;Z) →H^{k}(M;Z)## where the second arrow is the pull back under the zero section.

    - To define Chern classes as Euler classes one needs to show that every complex vector bundle is orientable and that one can choose an orientation in naturally determined way.
     
    Last edited: Nov 26, 2016
  14. Nov 27, 2016 #13

    lavinia

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    In general the transverse intersection of an orientable n-manifold with itself (viewed as the zero section of an oriented k-dimensional vector bundle) produces an orientable submanifold of dimension n-k. The homology class of this submanifold is Poincare dual to the Euler class of the vector bundle. This is a k-dimensional integer cohomology class.

    Chern has a paper in 1942 on a geometric proof of the Gauss Bonnet Theorem in all dimensions. He used a differential form that was an invariant polynomial in the curvature 2 form matrix of a Riemannian metric. This form generalizes the Gauss curvature times the volume element of a surface and represents the Euler class of the tangent bundle. I suspect that Chern classes came later - but not much.

    I am surprised that of all of Gauss's work the Gauss-Bonnet Theorem is less mentioned. It is the beginning of the theory of Characteristic Classes. In a sense, the theory began with invariant polynomials.
     
    Last edited: Nov 27, 2016
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