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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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    (Also: the fact that we have a Thom isomorphism is a consequence of the definition, it does not hold for an arbitrary cohomology theory.)2012-02-16

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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'!2012-02-20