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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 :) –  Mariano Suárez-Alvarez Dec 17 '10 at 15:11
    
@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). –  Akhil Mathew Dec 17 '10 at 15:28
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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. –  Sean Tilson Dec 17 '10 at 15:47
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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. –  Mariano Suárez-Alvarez Dec 17 '10 at 16:43
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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. –  Mariano Suárez-Alvarez Dec 17 '10 at 23:35

1 Answer 1

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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.) –  Akhil Mathew Dec 17 '10 at 17:28
    
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! –  Aaron Mazel-Gee Dec 17 '10 at 22:57
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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! –  Matt E Dec 18 '10 at 5:27
    
@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? –  Akhil Mathew Dec 18 '10 at 5:51
    
M&S use the Gysin Sequence (a consequence of the Thom isomorphism) and the euler class to construct the chern classes inductively. –  Sean Tilson Dec 18 '10 at 18:45

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