# Characteristic Class with arbitrary coefficient

I'm looking for some "natural definition" for characteristic class with arbitrary coefficient. For example, the chern class satisfies four conditions and Stiefel-Whitney class satisfies three conditions. Are there any axiomatic approach to define a characteristic class for Abelian groups $G$ other than $\mathbb{Z}$(Chern class) or $\mathbb{Z}/2\mathbb{Z}$(Stiefel-Whitney). Thanks!

• Every vector bundle (or even every principal G bundle) over a paracompact base is classified by a map from the base space to some classifying space, so there is a restriction map from the cohomology of the classifiyng space to the cohomology of the base space with whatever coefficients you like, and cohomology classes in the image of this map are called "characteristic classes". – James Cameron Jul 29 '14 at 14:37
• @JamesCameron Thanks for your explanation. But why did people use Chern class, Steifel Whitney class more often than classes with other coefficient? Does these classes contain more information? – Golbez Jul 29 '14 at 14:58
• I don't have a very good answer for that, other than that vector bundles are the first types of bundles that most people learn about, and $\mathbb{Z} /2$ coefficients work really nicely for arbitrary vector bundles since you don't have to worry about orientability, and $\mathbb{Z}$ coefficients work well with almost complex manifolds, where you usually see Chern classes. That's not a good answer though, I would love to learn more about specific types of principal bundles where taking characteristic classes with different coefficients is helpful. – James Cameron Jul 29 '14 at 15:55
• @JamesCameron Okay, thanks for your answer. This question occur to me accidentally, so I'm not sure if such example exists. By the way, I think the universal coefficient theorem could exchange the coefficient of Chern class, but there's no more extra information extracted from the exchange of coefficient. Also, the geometric meaning for the class after exchange of coefficient is not clear. So I'm searching for the so called "natrual definition". – Golbez Jul 30 '14 at 10:44
• It's Stiefel, not Steifel. – Georges Elencwajg Aug 2 '14 at 19:15

## 1 Answer

I don't think this is really what you had in mind, but here's another perspective on characteristic classes which lends itself to some generality.

Let $E,F$ be two multiplicative cohomology theories and $T \colon E \to F$ a natural transformation. Let $\Xi$ be some class of vector bundles which are naturally $E$- and $F$-oriented, in the following sense: each $\xi \in \Xi$, say of dimension $n$ over a space $X$, has Thom classes $u_\xi \in E^n(\xi, \xi-0)$ and $v_\xi \in F^n(\xi, \xi-0)$ which are natural in bundle maps, so that we have natural Thom isomorphisms $$\phi_\xi \colon E^*(X) \to E^{*+n}(\xi,\xi-0) \quad\text{and}\quad \psi_\xi \colon F^*(X) \to F^{*+n}(\xi,\xi-0),$$ given by multiplication by $u_\xi$ and $v_\xi$, respectively. Then we can define a characteristic class $$c_T(\xi) \mathrel{:=} (\psi_\xi^{-1} \circ T \circ \phi_\xi)(1) = T(u_\xi)/v_\xi \in F^*(X).$$ By naturality of the Thom classes and $T$ this will be natural in bundle maps. If we require our Thom classes to also be multiplicative, and $T$ to respect the multiplicative structures on $E$ and $F$, and the characteristic class will also be multiplicative.

Examples:

1. Take $E = F = \mathrm{H}\mathbb{Z}/2$, ordinary cohomology with $\mathbb{Z}/2$ coefficients, and $T = \operatorname{Sq} = \sum \operatorname{Sq^i}$ the total Steenrod square. All vector bundles are naturally $\mathrm{H}\mathbb{Z}/2$-oriented (this is the same thing as an orientation in the classical sense with $\mathbb{Z}/2$ coefficients). Then $c_{\operatorname{Sq}}$ is just the total Stiefel-Whitney class $w$ (this is immediate from the axiomatic definition of $w$), and taking the graded pieces we get $c_{\operatorname{Sq^i}} = w_i$.

2. I'm guessing the same story runs with $\mathbb{Z}$ coefficients to give Chern classes? But right now I don't know how cohomology operations work with $\mathbb{Z}$ coefficients so I'm not actually claiming this with any certainty; perhaps someone else can chime in. Note that these will only be defined for complex vector bundles because you need natural multiplicative $\mathrm{H}\mathbb{Z}$-orientations.

3. Take $E = F = \mathrm{K}$, complex K-theory, and $T = \psi^i$ an Adams operation. The resulting classes $c_{\psi^i}$ are called the cannibalistic classes, and were used by Adams to solve the vector fields on spheres problem (which is where I read about this stuff).

So I guess all I'm saying is that, from this perspective, characteristic classes in $\mathrm{H}R$, ordinary cohomology with coefficients in a ring $R$, come from Thom isomorphisms and natural transformations from some cohomology theory to $\mathrm{H}R$, e.g. cohomology operations on $\mathrm{H}R$.