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Is this question acceptable?: Why is the integral of the second chern class an integer

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    This is about http://mathoverflow.net/questions/59861/why-is-the-integral-of-the-second-chern-class-an-integer

    Why does this question live on? The person placing the question is either trolling, or beyond understanding an answer to his question.
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    My analysis of what is going on in the question:

    The OP seems to have a good understanding of de Rham cohomology, and basically no understanding of other approaches to cohomology. He thinks of H^k(X, Z) as the set of closed k-forms omega such that int_{sigma} omega is integral for every k-cycle sigma, module exact forms. (Note that this is not quite H^k(X, Z); it is actually H^k(X, Z) modulo torsion.)

    Now, for a line bundle, there is a very nice geometric picture of why (2 pi i)^{-1} int_{sigma} c_1(L) is integral for any sigma. It strikes me as a quite reasonable question to wonder how this picture generalizes. The question is asked in a confusing way, so that it wasn't clear at first which of two questions was being asked:

    (1) For a vector bundle V and a 4-cycle sigma, is there geometric argument for why (2 pi i)^{-2} int_{sigma} c_2(V) is integral? I would still like to know a direct answer to this.

    (2, the question which it is now clear that the OP meant) For a line bundle L and a 4-cycle sigma, is there a geometric argument for why (2 pi i)^{-2} int_{sigma} c_1(L)^2 is integral? As many posters have pointed out, this has nothing to do with line bundles. If you know about integral cohomology, the proof is simply that wedge product maps H^2(Z) x H^2(Z) to H^4(Z).

    If you only know about deRham cohomology, this strikes me as nontrivial! Given omega and eta closed 2-forms, such that int_{sigma} omega and int_{sigma} eta are integral for every 2-cycle sigma, why is it true that int_{tau} omega \wedge tau for every 4-cycle tau? I would like to know a direct answer to this.
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    I am going to assume that "Why is this question open?" is a figure of speech whose actual meaning is "This question does not deserve to be open." So I will skip my analysis of which questions get closed in practice, and skip to the merits.

    First of all, I use Wikipedia's definition: "A troll is someone who posts inflammatory, extraneous, or off-topic messages in an online community, such as an online discussion forum, chat room, or blog, with the primary intent of provoking other users into a desired emotional response or of otherwise disrupting normal on-topic discussion." I don't see how this is applicable, so I'll assume this isn't what you mean. By my definition, this clearly isn't trolling.

    Based on his other answers, the poster is clearly capable of understanding advanced math. Based on the vocabulary he uses in this question, I would say that he is probably reading physics-style sources, possibly ones which discuss Chern-Simons theory. And, as I say above, he only thinks in deRham cohomology, not integral cohomology. So, yeah, I think he definitely can learn from our answers. Do I think it belongs here? Well, it is at the level I want MO to be at -- material which is usually learned in grad school. I am especially sympathetic because I'm going to need to talk about the relation between integer and deRham cohomology soon in my class, and I am finding it tricky to get right, and because I never saw it in a formal course. But a lot of users seem to want a higher level and I don't have the energy to fight about it.
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    Ah, and I see the question is being continued over on math.SE http://math.stackexchange.com/questions/29797/direct-proof-that-the-wedge-product-preserves-integral-cohomology-classes Hopefully, that will make everyone happy.
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    My objection is not to the level of the question. My objection is the attitude of the poser.

    The title has nothing to do with what he is interested. He declares that his background is he undersands a book on De Rham cohomology (where the question he originally asked is answered). After taking some stabs, Donu Arapura answers it
    earnestly. The poser tells him I don't understand that stuff and don't care.

    The poser then names a new question, "Is the image of the integral cohomology in the DeRham cohomology a subring?" But wait, there is more, it must be done only using a definition that is completely detached from any knowledge of integral homology.

    Seems tough. Maybe it can be done, but I bet the proof would use ideas he doesn't know and doesn't care about.
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    @Charles: I can see where your objections are coming from, but after reading the question I do not myself feel that the OP is exhibiting an especially bad attitude. Like most of us, he knows some stuff but doesn't know some other stuff and is looking for an answer in terms of things he knows. He did thank someone for recommending Appendix C of Milnor and Stasheff, so presumably he will learn about the perspective of Donu's answer and will probably care more once he knows what Donu's talking about.

    To my mind, this person's greatest sin is calling himself "Greg Graviton". This does not encourage me to take him seriously...

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