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Hello,

'ordinary' Stiefel-Whitney classes are elements of the singular cohomology ring and are constructed using the Thom isomorphism and Steenrod squares. So I think they should exist for any (generalized) multiplicative cohomology theory for which the Thom isomorphism and cohomology operations like the Steenrod squares exist. If I am not wrong, I would be really happy about some references on this.

Thanks in advance (and merry christmas)

Jonas

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    $\begingroup$ Christmas is a very obstinate bachelor... $\endgroup$ Commented Dec 24, 2009 at 14:40

4 Answers 4

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Stiefel-Whitney classes exist for any real-oriented cohomology theory. This is a (multiplicative) cohomology theory E equipped with an isomorphism

$E^* ( \mathbb{R} P^{\infty} ) \cong E^*(pt) [[x]]$

The two most well known examples are ordinary cohomology (i.e. singular cohomology) with $\mathbb{Z}/2$-coefficients and also unoriented bordism theory MO (this is the Thom spectrum MO, not "MathOverflow"). Note that such an isomorphism might not exist. For example it doesn't exist for ordinary Z cohomology, nor does it exist for K-theory.

The choice of this generator x is the first Stiefel-Whitney class and the other classes can be constructed using the splitting principle. For such a theory you will have Thom isomorphisms for all real bundles (not necessarily oriented).

I believe this is all explained in Switzer's book "Algebraic Topology", but I'm not sure. The course I learned it from didn't have a text book. This is often standard material for a second semester of graduate level algebraic topology, at least it was at UC Berkeley.

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  • $\begingroup$ Thanks, I'll have a look at that book as soon as my library opens again.. $\endgroup$ Commented Dec 24, 2009 at 16:20
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    $\begingroup$ Yes, this is indeed in Switzer, around 16.29. $\endgroup$ Commented Dec 24, 2009 at 17:45
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A well-studied example of this are the "cannibalistic characteristic classes" of Adams and Bott. If I remember correctly, there is a nice account of this in Bott's "Lectures on K(X)"; they are also discussed in Adams' J(X) papers (in Topology in the mid 60s).

The cannibalistic classes $\theta_k(V)$ of a complex bundle $V\to X$ are defined by the formula $$\psi^k(u) = \theta_k(V)\cdot u$$ where $\psi^k$ is an Adams operation, and $u$ is the standard Thom class in $K(\mathrm{Th}(V))$ (K-theory of the thom space of $V$). Thus, the role of Steenrod operations is played by the Adams operations.

The cannibalistic classes can be used to detect when two bundles $V$ and $V'$ are not "stably fiber homotopy equivalent"; i.e., when the sphere bundles $S(V\oplus \mathbb{R}^m)$ and $S(V'\oplus \mathbb{R}^m)$ are not fiber homotopy equivalent for all $m$. (See either of the references for this.)

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    $\begingroup$ Why "cannibalistic"? $\endgroup$
    – Qfwfq
    Commented Apr 19, 2010 at 7:31
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    $\begingroup$ Frank Adams' peculiar sense of humor. The idea is that K-theory is feeding on itself to create new characteristic classes. $\endgroup$
    – Peter May
    Commented Apr 26, 2012 at 12:36
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Maybe this isn't the "right" way to think about them, but I have often found the following characterization much more appealing: the characteristic classes are just elements of $E^*(BG)$ for some cohomology theory and some group $G$, if $E$ is ordinary cohomology with integer coefficients and $G$ is the infinite unitary group you get Chern classes etc. The universal property of $BG$ gives you a unique map that classifies each $G$-bundle, now look at the map in cohomology that you get from this and you get the characteristic class of the desired bundle. This probably requires some sort of orientability for things to be "nice" but you can certainly define characteristic classes for any cohomology theory this way, they just might not have nice relations.

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    $\begingroup$ Put another way: E characteristic classes for G -bundles are natural transformations {G-bundles/X} ---> E^*(X) Since {G-bundles/X} is $[X,BG]$ the Yoneda lemma tells you such natural transformations are in one-to-one correspondence with $E^*(BG)$. $\endgroup$ Commented Aug 18, 2010 at 23:20
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Not a real answer to your question, but I think it may be related. One can ask the same question fofr Chern classes. For some generalized cohomology theories you can define Chern classes of complex vector bundles, and these satisfy the usual axioms for Chern classes.

What changes is the tensor product behviour. Given two line bundles L and M, the first Chern class $c_1(L \otimes M)$ is given by a universal power series in $c_1(L)$ and $c_1(M)$. For instance in ordinary cohomology $c_1(L \otimes M) = c_1(L) + c_1(M)$, while in K theory $c_1(L \otimes M) = c_1(L) \cdot c_1(M)$. Since line bundles form a group under tensor product, this power series is a (1-dimensional) formal group law.

So to any such cohomology theory you can attach a 1-dimensional formal group. For instance you attach the additive group to ordinary cohomology and the multiplicative group to K-theory. It turns out that you can go the other way round. The other 1-dimensional formal group laws are formal expansions of the group law of an elliptic curve near the origin; these give rise to the so called elliptic cohomology theories.

I think you can find details on the above constructions in any reference about elliptic cohomology. I've only heard about these theories, so I don't know an good reference. By the way, since I'm not an expert, please correct me if anything I have written above is wrong.

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    $\begingroup$ In K-theory you have $c_1(L\otimes M) = c_1(L) + c_1(M) - v c_1(L) c_1(M)$ where $v \in K^{−2}=K_2$ is the Bott class. This is the multiplicative group in the sense that $(1- v c_1(L \otimes M)) = (1 - v c_1(L)) (1 - v c_1(M))$ There are other normalizations: $1+ v c_1(L)$ gives all plus signs and in periodic K-theory we could just work in $K^0$, or equiv, set $v=1$. The form I wrote has the virtue of linking Chern classes in cohomology, connective K-theory, and periodic K-theory: the Chern classes in connective K-theory specialize to those in ordinary cohomology and in periodic K-theory. $\endgroup$ Commented Aug 18, 2010 at 23:17

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