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Recall that complex topological $K$-theory is representable on reasonable spaces by the space $BU \times \mathbb{Z}$ (where $BU$ is a colimit of various infinite Grassmannians), and that the total Chern class provides a natural map $\mathrm{Vect}(B) \to H^*(B)$ for every such space $B$. By the multiplicativity property, this map factors through the K-group and leads to a natural transformation $K(B) \to H^*(B)$. $H^*(B)$ is also representable by a product of Eilenberg-MacLane spaces $K(\mathbb{Z}, n)$ over all $n$. There is thus a map, unique up to homotopy $$BU \times\mathbb{Z} \to \prod_n K(\mathbb{Z},n).$$

What is this map?

As Mariano observes, one can simply define the individual Chern classes on the $K$-group as well, albeit not immediately through the universal property, so the question reduces to the determination of the (homotopy class of the) map $BU \times \mathbb{Z} \to K(\mathbb{Z},n)$ induced by each Chern class.

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    It will be the product of all maps $BU\times\mathbb Z\to K(\mathbb Z,n)$ representing the individual Chern classes, so there is little gain in considering the whole product :)2010-12-17
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    @Mariano: OK, sure. I initially considered the product only because it wasn't obvious to me that individual Chern classes work on the level of the K-group (but the total Chern class clearly does, so the individual ones have to.. .fair enough).2010-12-17
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    I am a little confused. Are you asking essentially for an element of $H^*(BU;\mathbb{Z})$? this we know. They are the universal chern classes right? I guess you could say that the map is the universal chern class, but I am not sure if this is what you are after.2010-12-17
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    Just to pursue Sean's remark: my understanding is that one can compute the cohomology ring of $BU$ more or less directly, and so (as Sean writes) read off the universal Chern classes. These are then classified by the maps to $K(\mathbb Z,n)$ that you are asking about. Is the point that you want to describe these maps more explicitly than via this two-step process? Do you have any reason to think that there is a particularly natural description? (Not that I have any reason to be sceptical, I'm just curious.) Also, what model for $K(\mathbb Z,n)$ do you have in mind? Without something ...2010-12-17
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    ... reasonably concrete, I'm not sure how you're going to do better than to say "the map induced by this cohomology class in degree $n$", which will put you back into the situation of Sean's comment.2010-12-17
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    @Sean: Right, OK. This makes sense, and I would be happy to accept that as an answer if you posted. @Matt E: I was in fact curious if there was an explicit description of the map as a homotopy class, but you're of course right that beyond $K(\mathbb{Z}, 2)$, there isn't much of an explicit description. For some reason I applied the fact that $K(\mathbb{Z}, n)$ classify cohomology at every step but the last, when I forgot about it, and with that -- as Sean observes -- the question is straightforward.2010-12-17
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    Dear @Matt E: Is there a straightforward way of obtaining $H^*(BU)$? I know that an inductive argument with spectral sequences will do it, and I'm pretty sure Schubert calculus can also do it, but it's somewhat messy and non-elementary.2010-12-17
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    Well, there are reasonably nice models for $BU$ and for $K(\mathbb Z,n)$, so one can wish for a nice description of the actual maps between those models. math.stanford.edu/~ralph/charact.ps seems to do precisely that in the algebraic case.2010-12-17
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    @Akhil: I don't understand how Sean's answer answers the question! I though you were asking, in essence "Give me a model of $BU$, a model of $K(\mathbb Z, n)$, and a map between them which represents the $n$th Chern class. Sean's nice answer does not answer *that*. If you follow the link in my earlier comment, you'll see it done in the category of algebraic varieties: howpefully someomne can give us the details in the case of spaces, which should be considerably easier.2010-12-17
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    @Mariano: Thanks! I saw the paper, though while it looks interesting, I'm not currently advanced enough in homotopy theory to have anything much to say about it (or to understand much of what's going on).2010-12-18
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    @Mariano: Dear Mariano, Thanks for your comment and link, which directly address my initial question (in a positive direction, which is much better than the other direction!).2010-12-18
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    @Mariano: I wasn't sure how I answered Akhil's question either. Part of the point is, from the perspective of homotopy theory, we don't really need a model of the map Akhil is asking for. In fact, I don't know if that model helps me do anything. I could be completely wrong though.2010-12-19

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One way, my favorite, of defining the chern classes goes as follows: first compute $H^*(BU; \mathbb{Z})=\mathbb{Z}[c_1, c_2, ...]$ by using your favorite method. I know two methods, one is using some cellular description of the grassmanians (see Milnor and Stasheff), the other is to first compute the cohomology of the lie groups $U(n)$ and then use a path-loop space fibration and the Serre SS to get at the cohomology of $H^*(BU(n);\mathbb{Z})$ (see Homology and Euler characteristics of the classical Lie groups). Next you notice that $BU$ and $BU(n)$ have the same cell structure through a range and so you essentially take the limit (colimit, and there are some subtleties here, the place I recall seeing these is in Jacob Lurie's survey on elliptic cohomology, at the beginning).

Next suppose you have a complex vector bundle $\xi : E \to B$ then it is classified by a map $f: B \to BU(n)$ where n is the dimension of $\xi$. You can get a map on all of $BU$ by just adding on trivial bundles to get a map out of $BU(k)$ for all $k$ larger than $n$. This gives a map out of BU (although technically you don't really need this, $BU(n)$ works fine for defining the chern classes of an $n$ dimensional complex vector bundle). Now define $c_n(\xi):=f^*(c_n)$.

From this perspective we started with the universal case, so maybe it is a bit of a cheat. For me this even clarifies how I should think about characteristic classes in general. Suppose you want to look at $G$-bundles and see what you can tell about them from $E$-theory ($E$ some ring spectrum and $G$ some compact lie group or whatever you need for $BG$ to be nice, I am not sure if there are other restrictions for this to work). Now compute $E^*BG$ and use the fact that $G$-bundles over $X$ are classified by homotopy classes $X \to BG$. You should check out some of the threads on MO about characteristic classes, I think I can learn something from each of Rezk's answers.

please let me know if I can make some of the above clearer or if there are any mistakes.

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    Thanks! (And extra characters.)2010-12-17
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    I agree that this is definitely the best way of constructing Chern classes, it's so much nicer than talking about curvature forms and all that. One time we spent a whole lecture in my symplectic geometry class trying to show why $c_1(L)\in H^2(X;\mathbb{Z})$ classified line bundles!2010-12-17
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    @Aaron: Dear Aaron, I just want to remark that there are lots of reasons for considering the definition of Chern classes via curvature (or via obstruction theory, or via $K$-theory, or via lots of other points of view). So while I agree that the picture in terms of cohomology of the classifying space is very beautiful and satisfying, it's good to be open-minded about other points of view too!2010-12-18
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    @Aaron: Milnor-Stasheff construct the Stiefel-Whitney classes via Steenrod squaring the fundamental class of the bundle and applying the Thom isomorphism. I guess that doesn't work for Chern classes anymore since they're over $\mathbb{Z}$, but is there another way via different cohomology operations?2010-12-18
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    M&S use the Gysin Sequence (a consequence of the Thom isomorphism) and the euler class to construct the chern classes inductively.2010-12-18
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    Perhaps morally there should be a similar construction though? I have no idea what $H\mathbb{Z}^*H\mathbb{Z}$ is, though I suppose one could try to find it by Bocksteining $H\mathbb{F}_2^*H\mathbb{Z}$...2010-12-27
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    @ Matt E: Certainly! It's just that we had even talked about how $\mathbb{CP}^\infty=K(\mathbb{Z},2)$, so it was pretty silly that we didn't bridge the gap there.2010-12-27
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    @Aaron: that would only give you the 2-local part of $H\mathbb{Z}^*H\mathbb{Z}$. So you would have to do that for each prime and then assemble them. I am pretty sure no one currently has the full computation done.2010-12-27
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    Right, of course. So I guess $H\mathbb{F}_p^*H\mathbb{Z}$ must vary substantially as we change $p$ -- or maybe $p=2$ is just different from $p\not= 2$, as I've heard is often the case with such things...2010-12-27
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    well, sort of. Certainly for odd primes the algebra is just a bit different, we can see that much just from the axioms. But take a look at one of the constructions of the operations, the hands on ones not the homotopical one, and I just have no clue what how one could go about constructing an integral operation. I just don't get it. maybe some version of a bockstein I could understand, but other than that I am at a loss. I think that this issue is in some sense comparable to solving diophantine equations at every prime and then trying to get an integral solution.2010-12-27
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    For reference for anyone who cares, Sean posed this on MO and got some reasonable answers: http://mathoverflow.net/questions/50519/integral-cohomology-stable-operations2010-12-30
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    @Aaron I was shown a "definition" of the chern classes using the looking at some eigenvalues of some operator related to the curvature form. That was amazing after all this time, I am always thrilled to see topology have an impact on geometry like that. Just like what happens in Morse theory, so I think the benefit of that construction is not that it provides a definition, but that it relates two fields. I could be wrong though, and I guess you can actually compute things with that definition though, although that seems much harder to me...2011-08-08