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Why does Chern class belong to INTEGER cohomology class?

  1. Aug 18, 2010 #1
    Hi all,

    I am new to topology & geometry. I skimmed through the subject in various books so I might be asking simple questions.

    My question is how can we say Chern's class belong to Integer rather than Real cohomology class? I assume one defines it through the characteristic polynomial of the curvature form. (as in http://en.wikipedia.org/wiki/Chern_class) It does not look obvious to me this make chern's class belong to integral cohomology. I think this is important so that Chern's numbers are integers.
     
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  3. Aug 18, 2010 #2

    lavinia

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    I have never seen a direct proof that the integrals of products of Chern forms are integers except in the case of the Euler class of the tangent bundle which is the Gauss Bonnet theorem. I would love to see such a proof.

    The approach I have seen is that the Chern forms of induced complex vector bundles with induced connection are the pull backs of the Chern forms of the image bundle. Since one also proves that the cohomology class of a Chern form is independent of the connection (and is also a closed form) one gets natural cohomology classes of complex vector bundles.

    One also proves the product formula for the Whitney sum of complex vector bundles.

    These properties by themselves characterize the Chern classes uniquely. Naturality implies that Chern classes are pull backs of the Chern classes of the universal bundle in the cohomology of the universal classifying space for complex vector bundles. Thus one only needs to know that the Chern forms are the images of integer classes in the cohomology of the universal classifying space under the coefficient sequence 0 -> Z -> R ->R/Z -> 1
     
  4. Aug 18, 2010 #3
    I don't know most of those Mathematical terms. Can you give me some references? I have the book by Nakahara but it doesn't seem to mention Chern number directly.
     
  5. Aug 18, 2010 #4

    lavinia

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    ask some questions. I will answer them.
     
  6. Aug 20, 2010 #5

    lavinia

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    The references are advanced and cover a lot of material. Here is a sketch.

    Chern forms exist for complex vector bundles.

    One needs to understand what these bundles are and how you can obtain new bundles from old ones.

    The most important way to create a new bunde is the induced bundle of a continuous map.

    if E is a vector bundle over M and f:N ->M is a continuous map then the subset of NxE of points (x,(f(x),v)) is the induced bundle.

    If the map is smooth then a connection on E pulls back to a connection on the induced bundle. As a result the curvature forms of the connection on E pulls back to the curvature forms of the connection on the induced bundle. One shows easily that f* Chern form on E is the corresponding Chern form on the induced bundle.


    Now all bundles are induced from some map into a universal classifying space. What this means incredibly is that there is a bundle over some space so that all bundles are induced from this bundle by some map into this space. This "universal bundle" also has a connection and so has Chern forms derived from curvature. All Chern forms are thus pull backs of these.

    So the statement that Chern forms are really integer cohomology classes reduces to the special case of the universal bundle.

    A few other things need to be proved. First that the Chern forms are closed and thus represent cohomology classes. This is easy.
    Second one needs to show that the cohomology class is independent of the connection.

    Once this has been done one know that the Chern forms are cohomology classes that are characteristic of the bundle independent of its geometry.

    Also note that a connection on a bundle does not require a metric. Connections and curvature are more fundamental than metrics and indeed there are connections that are not compatible with any metric.

    For real vector bundles the analogues of Chern forms are all zero. They only exist for complex vector bundles. But one can complexify a real vector bundle ( tensor each fiber with the complex numbers) and take the Chern classes of the complexified bundle. These are called the Pontryagin classes of the bundle. One easily shows that the Pontryagin classes are non-zero only in dimensions that are a multiple of 4.
     
    Last edited: Aug 20, 2010
  7. Sep 20, 2010 #6

    mathwonk

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    I think it helps to know something about what they mean. For complex holomorphic line bundles on manifolds for instance, the first chern class seems to be Poincare dual to the integral homology class of a section. Hence on a Riemann surface the 1st chern number is the number of zeroes of a holomorphic section.

    In sheaf terminology the exponential map takes the sheaf O of holomorphic functions to the sheaf O* of never zero holomorphic functions, with kernel Z the sheaf of integer valued functions. The chern class map is the connecting homomorphism from the group H^1(O*) classifying line bundles to the group H^2(Z), hence has integer classes as values.

    This is ultimately the fact that the logarithm function has an integer obstruction to being well defined as you go around a loop.

    In general as Lavinia says, chern classes are always pull backs of the classes of the tautological bundle on a grassmannian, so it suffices to know they are integral in that one example.

    Also one could look in milnor and stasheff, p. 158, where they are defined inductively in terms of integral euler classes. basically all of them can be defined in terms of the top degree ones which are integral euler classes. this is the example above for line bundles where the top class is also the first one.

    another approach to reducing to the case of line bundles is to use the splitting principle by which any bundle is the image of a sum of line bundles, hence its chern classes map injectively to products of classes of line bundles.
     
    Last edited: Sep 20, 2010
  8. Sep 20, 2010 #7

    mathwonk

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    another approach is to use the characterization in bott tu of the chern classes via the leray hirsch theorem which expresses the cohomology of a fiber space in terms of a polynomial in the cohomology of the fiber with coefficients in the cohomology of the base. then the fact that the leray hirsch theorem holds over the integers, as in hatcher's book, should give you the fact that these coefficients, i.e. the chern classes of the vector bundle, are integer classes.
     
  9. Sep 20, 2010 #8

    lavinia

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    The top dimensional Chern class is the Euler class of the bundle.
     
  10. Sep 20, 2010 #9

    mathwonk

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    is there any difference here? on a line bundle i think the top class is the first class.

    maybe i should have said dual to the homology class of the zero locus of a section.
     
  11. Sep 21, 2010 #10

    lavinia

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    I wasn't arguing with you at all. I was just observing that your example of the first Chern class of a complex line bundle over a Riemann surface illustrates an Euler class. In general, the intersection manifold of the self intersection of the manifold in the bundle is Poincare dual to the Euler class. When the bundle is a n plane bundle over an oriented n manifold this Poincare dual is just the sum of the zeros of the intersection counted with orientation and multiplicity.

    I guess technically, the top Chern class of the complex vector bundle is the Euler class of the underlying real vector bundle.
     
    Last edited: Sep 21, 2010
  12. Sep 21, 2010 #11

    mathwonk

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    Yes I see now you were generalizing my claim. Let me see if I understand this. so you are assuming oriented manifold I guess, and perhaps also compact (as I was) in order to define self intersection and orientation of zeroes, and have finiteness of zeroes. it seems a little interesting to define self intersection still, since the bundle itself is not compact. (I think some people use the diagonal inside the product of the manifold with itself to do this self intersection, perhaps because that manifold is compact, at least in the case of the tangent bundle, but maybe you can always intersect compact submanifolds of a manifold?) and I guess there is no unique intersection manifold, so you are using some transversality argument to say there exists a smooth representative of this homology class. And by zeroes of the intersection I guess that means oriented points of intersection, unless of course you are thinking of the manifold as the locus of zeroes of the fibers of the bundle. So the points of intersection represent the zeroes of a section of the bundle in my terminology.

    Of course holomorphic sections do not always exist, whereas your self intersection does make sense in the smooth sense. It seems interesting that the top chern class of a complex holomorphic bundle does not depend on the holomorphic structure. So apparently the chern classes do not use the holomorphic structure at all, just the smooth structure? I am so used to holomorphic objects. This did not occur to me so clearly until your remark that the euler class depends only on the underlying real manifold. Ah yes, the complex structure determines the orientation. Is that all it is used for?
     
    Last edited: Sep 21, 2010
  13. Sep 21, 2010 #12
    The top chern class is the euler class, so this does not depend on the holomorphic structure, just the orientation. For lower Chern classes, this is no longer the case.

    The proof that the curvature of a connection is in fact an integral cohomology class depends on the splitting principle and the Gauss-Bonnet theorem for the 2-sphere.
     
  14. Sep 21, 2010 #13

    lavinia

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    Every oriented vector bundle has an Euler class. Every complex vector bundle is canonically orientable.
    Every holomorphic section of a complex line bundle over a Riemann surface must have isolated zeros - I think.

    I think this is the two ways that holomorphic comes in.
     
  15. Sep 21, 2010 #14

    lavinia

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    I apologize for being vague. Here is a little more detail. I am pretty sure that this is right but... correct me if if I am confused.

    The Thom class of an oriented k-plane bundle over a smooth n-manifold is Poincare dual to the zero section of the bundle. ( Bott and Tu Differential Forms in Algebraic Topology pgs 66-67)


    The Euler class of the bundle is the pull back of the Thom class under the inclusion map of the manifold (Bredon Topology and Geometry pg 378 or Milnor's Characterisitc classes chapter on the Euler class) ) in the bundle as the zero section. If one takes a map that is homotopic to this inclusion but is transversal to the zero section the one gets an n-k dimensional submanifold from the intersection of the manifold with its image under this transversal mapping. This is what I meant by self-intersection.

    This n-k dimensional manifold when viewed as a submanifold of M is Poincare dual to the pull back of the Thom class i.e. to the Euler class ( Bott and Tu pg 69).

    In the case of the tangent bundle or any other n-plane bundle the n-k dimensional manifold is zero dimensional and is thus a finite collection of points (if M is compact). These points are the isolated zeros of the section and I am pretty sure that they must all have multiplicity 1 or -1.
     
    Last edited: Sep 21, 2010
  16. Sep 21, 2010 #15

    mathwonk

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    Ok thanks. I have another question now generated by your other remarks. Is the chern class actually a topological invariant and not just a smooth invariant? I.e. we have given several definitions that use smoothness, via either transversality, or curvature forms. But i wonder if changing the smooth structure of the manifold changes the chern classes? I guess not. Could we argue this by using the topological leray hirsch theorem? I.e. I suppose the leray hirsch theorem holds topologically, and expresses the cohomology of the bundle as a polynomial ring over the cohomology of the base space. then the "characteristic" polynomial giving a relation for the relative first chern class of the fundamental line bundle of the projectivized bundle should define the chern classes as coefficients of this polynomial.

    have you thought about this?
     
  17. Sep 22, 2010 #16

    lavinia

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    Don't know but the question is great and has me reading.

    The only thing that I do know is that the statement of Novikov's theorem which is that the rational Pontryagin classes of two homeomorphic smooth manifolds are equal. I guess this means that the integer classes may differ by torsion elements.

    I will keep reading.

    One question just to make sure I understand what you are asking. Are you only asking about the Chern classes of the tangent bundle? If so then this is a special type of manifold - not sure exactly - but a manifold whose tangent bundle admits a complex structure.
     
  18. Sep 22, 2010 #17

    lavinia

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    I guess a related question is whether a given smooth manifold can have two almost complex structures with different Chern classes. Then the Chern classes would not even be smooth invariants.

    According to the introduction to the paper in this link,

    http://math.gcsu.edu/~ryan/research/exotic/invariants.pdf [Broken]

    there are examples of almost complex structures on the complex projective 3 space (considered as a smooth manifold) that have different total Chern classes.

    If I understand this right - it implies a point that I had never thought of - that is that a connection on a complex vector bundle must depend specifically on the almost complex structure - otherwise the Chern classes would have to be the same for all almost complex structures.
     
    Last edited by a moderator: May 4, 2017
  19. Sep 22, 2010 #18

    mathwonk

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    yes i realize my question was not clear, as it was not clear in my mind. i keep confusing classes of given bundles with classes of manifolds, based on their tangent bundles. so I guess if we use the definition of chern classes via the pullback from the universal classes on the grassmannian, we should get that a continuous equivalence of bundles implies they have the same classes.

    but if we are only given two smooth manifolds which are homeomorphic, we have to ask if their tangent bundles yield the same classes, since I do not know whether the tangent bundles of two smooth but homeomorphic manifolds are continuously equivalent.

    And if we only have a continuous manifold, possibly with no smooth structure?, and try to ask how to define its chern or other classes of this topological manifold, we seem to need a definition of a "topological tangent bundle".

    I found a survey article in the American Math Monthly by Richard Lashof on this in december 1972, but i do not have reading access at the moment.
     
  20. Sep 22, 2010 #19

    lavinia

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    The example I found above for complex projective 3 space is of different almost complex structures on the tangent bundle that have different Chern classes. (The underlying real bundle does not have Chern classes - only Pontrjagen classes.)

    Two homeomorphic smooth manifolds must have the same rational Pontrjagen classes but I suppose they can differ by torsion elements. I would like to see an example.
     
  21. Sep 22, 2010 #20

    mathwonk

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    I found that link very interesting indeed. I did not even know that CP^3 was diffeomorphic to S^6 blown up at one point. although maybe I should have. I.e. blowing up is topologically the same as taking a connected sum with what? C^3?

    So the connected sum just tacks on a disc to a hole in CP^3, which does nothing to CP^3 I guess? or viewed the other way, it tacks on a CP^3 to a punctured S^6 which is a disc, so same difference I guess.

    I guess what I am trying to say is that taking a connected sum with a sphere should not do anything. So conversely taking a connected sum of a sphere with X should still be X.

    hence blowing up an (almost complex) sphere at a point should be topologically a complex projective space.

    but i have never done the details of this myself. Also it seems orientation gets messed up. For one thing CP^3 does not have any holomorphic subvarieties that can be blown down, so it is not even I guess almost holomorphically the same as a blown up object.

    or rather not the same as a blown up holomorphic object. of course that is the open problem, whether S^6 can be viewed as a holomorphic object.

    mumbkle mumble, very interesting examples. thank you.
     
  22. Sep 22, 2010 #21

    mathwonk

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    so your example either proves that chern classes are not smooth invariants, or that the two bundles are not smoothly equivalent.

    so i guess you are saying it proves they are not smooth invariants? yes it seems so.
     
  23. Sep 22, 2010 #22

    mathwonk

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    that interesting paper seems to be at least derived from brown's 2006 phd thesis with segert at mizzou, in columbia, mo.

    are you a topologist? i always wanted to be a differential topologist but ultimately went in a different direction (algebraic geometry). It seems complex algebraic geometry yields a lot of interesting examples in topology though, and not just for 4 manifolds.
     
  24. Sep 22, 2010 #23

    mathwonk

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    do you think the pontryagin classes of homeomorphic smooth manifolds can actually differ by a non zero torsion element?

    say does your example show that the chern class is not a smooth invariant of a complex manifold, or is it key that the almost complex structure is not integrable?

    I guess I thought the cherm classes of a complex projective variety were smooth invariants. maybe the kahlerness is involved, as perhaps mentioned in that paper?
     
  25. Sep 22, 2010 #24

    lavinia

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    I do not think it proves that and this is really out of my area of knowlege. I think that the question you are asking is whether a smooth manifold can be a complex manifold in two inequivalent ways - i.e. there is no biholomorphic diffeomorphism between the two - and have different total Chern class. let's research it.

    I suspect that there must be examples of homeomorphic smooth manifolds with different Pontrjagen classes just because I have never heard an improvement on Novikov's theorem - but I am not sure.
     
  26. Sep 22, 2010 #25

    lavinia

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    BTW: The idea of tangent bundle has analogues outside of the smooth category - in the piecewise linear category and in the topological category. I know nothing of this theory but do know that there are triangulated manifolds whose triangulation can no be smoothed i.e. there is no differentiable structure on the manifold that is compatible with the triangulation. One way to find such a manifold is to calculate is combinatorial Pontrjagen numbers and show that they are not integers. In Milnor's characteristic classes he constructs an triangulated 8 manifold that has no compatible smoothness structure.
     
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