# The definition of a complex oriented cohomology theory

Here are two definitions for complex oriented cohomology theories.

A complex oriented cohomology theory $E$ is a multiplicative cohomology theory

• With a class $x \in \tilde{E}^2(\mathbb{C} P^\infty)$ whose restriction under the composite $$\tilde{E}^2(\mathbb{C} P^\infty) \to \tilde{E}^2(\mathbb{C} P^1)=\tilde{E}^2(S^2) \simeq \tilde{E}^0(\ast)$$ is 1
• That has a choice of Thom class for every complex vector bundle. That is if $\xi \to X$ is a complex vector bundle of dimension $n$ then we are given a class $U = U_\xi \in \tilde{E^{2n}}(X^\xi)$ with the following properties:
• For each $x \in X$ the image of $U_\xi$ under the composition $$\tilde{E}^{2n}(X^\xi) \to \tilde{E}^{2n}(\ast^\xi) \to \tilde{E}^{2n}(S^{2n}) \simeq E^0(\ast)$$ is the cannonical element 1
• The classes should be natural under pullback: if $f:Y \to X$ then $U_{f^*\xi} = f^\ast(U_\xi)$
• $U_{\xi \oplus \eta} = U_\xi \cdot U_\eta$

I have seen both used in the literature. For example Ravenel takes the first, whilst in the COCTALOS course notes, Hopkins takes the second, and states the equivalence with the first as Proposition 1.3 (which is never proved).

How exactly does one show these are equivalent? The only thing I can really think of is to somehow use the Thom isomorphism $E^*(X) \simeq \tilde{E}^{\ast+2n}(X^\xi)$

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Maybe it's helpful to know that $CP^{\infty}$ classifies complex line bundles... and do you know about the splitting principle? –  Dylan Wilson Feb 16 '12 at 8:53
@Dylan - Thanks, I am aware of both these. Are you saying to apply this to the bundle $\xi \to X$? i.e. then we have the flag variety $\text{Fl}(\xi)$ with a map $p:\text{Fl}(\xi) \to X$ such that the pullback bundle splits as a direct sum of line bundles and $E^*(\text{Fl}(\xi)) \to E^*(X)$ is injective? –  Juan S Feb 16 '12 at 10:22
Actually, I guess I have to be a bit more careful: The splitting principle holds for ordinary cohomology, but to prove it works for a complex-oriented theory, we need to prove it. Basically this follows from the fact that the distinguished element in E^2CP^{\infty} gives a firct Chern class, and we can prove a sort of "projective bundle formula" for it. I'll write some details later. –  Dylan Wilson Feb 16 '12 at 17:06
(Also: the fact that we have a Thom isomorphism is a consequence of the definition, it does not hold for an arbitrary cohomology theory.) –  Dylan Wilson Feb 16 '12 at 17:06

## 1 Answer

I was very confused about the seeming disparity between definitions as well. I would recommend you take a look at Adams blue book, he shows how using the definition Ravenel uses you can compute quite a lot about the homotopy of complex grassmanians. From there you can try and prove a thom isomorphism in the universal case (maybe he even does this). Also, try thinking about what an Atiyah-Hirzebruch-Serre Spectral sequence might be. Do you know the proof of the Thom isomorphism using the Serre Spectral Sequence?

You really should look at Adams blue book, part 2.

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Thanks for reminding me to check Adams' book. I read part 3 a while ago and then never really got into Part 2; I think I was discourage by his claims that everything was 'clear' and 'obvious'! –  Juan S Feb 20 '12 at 22:47