complex-oriented cohomology theories

Question

Assume that $E$ is a (reduced) extraordinary multiplicative cohomology theory. Then it is called complex-oriented if there is a class $c^E_1 \in E^2(\mathbb{C}P^{\infty})$, such that the map $i^*: E^2(\mathbb{C}P^{\infty}) \to E^2(\mathbb{C}P^{1}) \cong E^*(pt.)$, induced by the inclusion $i : \mathbb{C}P^{1} \to \mathbb{C}P^{\infty}$, sends $i^*(c^E_1)=1 \in E^*(pt.)$. However, this probably is the short story of the real meaning behind the complex-oriented idea. I found somewhere that this definition, implies the existence of a Thom class for $E$. In other words, for a complex orientation of $E$ suffices to specify a Thom class and vice versa. How do we prove that these two facts are equivalent?

Answer 1

The statement to show is:

$E$ is complex-oriented iff there is a Thom class $u_V \in E^{2k}(X^V)$ for any complex vector bundle of rank $k$.

The $(\Leftarrow)$ direction follows from choosing the tautological line bundle $L$ over $\mathbb{C}P^\infty$. We get a Thom class $u_L \in E^2((\mathbb{C}P^\infty)^L) \cong E^2(\mathbb{C}P^\infty)$ since $(\mathbb{C}P^\infty)^L \simeq \mathbb{C}P^\infty$. By definition/by the Thom isomorphism, the restriction of the Thom class $u_L$ to a fiber gives the generator of $E^2(S^2)$.

For the $(\Rightarrow)$ direction, if $V$ is a line bundle, we are done by pulling back the first Chern class $c_1^E$ along the map on Thom complexes induced classifying map for $V$. We can reduce the general case to the rank 1 case by the splitting principle.

Here are some more details. Suppose $V$ is a line bundle over $X$. There is a pullback of line bundles \begin{array}{ccc} V & \rightarrow & L \\ \downarrow & & \downarrow \\ X & \rightarrow & \mathbb{C}P^\infty. \end{array}

There is an induced map on Thom complexes $X^V \to (\mathbb{C}P^\infty)^L \simeq \mathbb{C}P^\infty$. Since $E$ is complex-oriented, there is a first Chern class $c_1^E \in E^2(\mathbb{C}P^\infty)$, which we can then pullback along this map to get an element $u_V \in E^2(X^V)$. To show that $u_V$ is indeed a Thom class, we observe that it's true by complex-orientation $c_1^E$ begets a generator when you restrict it to a fiber of the Thom complex (which is $S^2$). Then we use the above pullback square to deduce that the fibers of the Thom complexes are identified with each other, so $u_V$ also restricts to a generator along a fiber.

Now suppose $V \downarrow X$ is an arbitrary complex vector bundle of rank $k$, and the argument is to use the splitting principle to reduce this to the previous case. As you say, there is a little bit of work needed to extend this to complex-oriented cohomology, but the argument is essentially the same. Recall how the splitting principle for ordinary cohomology is proved. We need to show:

1. Pulling back $V$ along the map $\mathbb{P}V \to X$ yields a vector bundle that splits off a line bundle.
2. The induced map $E^*(X) \to E^*(\mathbb{P}V)$ is monic.

(1) is just geometry, and nothing changes from the original case. (2) is more interesting. We claim a fortiori that $E^* \mathbb{P}V \cong E^* X [t]/c_t(V)$. This follows from the Leray-Hirsch theorem for $E$-cohomology, and the essential ingredient of that theorem is the collapse of the Atiyah-Hirzebruch-Serre spectral sequence for $\mathbb{C}P^{k-1} \to \mathbb{P}V \to X$. Using complex-orientability again, we see that the element $c_1^E \in E^2 \mathbb{C}P^{k-1}$ in the spectral sequence must be a permanent cycle, and using the ring structure we see that all of its powers are permanent cycles too. This verifies the condition of the Leray-Hirsch theorem, and the claim is proved.

Returning to the main argument, we can repeatedly apply the splitting principle to get some pullback \begin{array}{ccc} L_1 \oplus \cdots \oplus L_k & \rightarrow & V \\ \downarrow & & \downarrow \\ F & \rightarrow & X. \end{array}

We also get Thom classes $u_{L_i} \in E^2(F^L_i)$. Using the external cup product and $F^{L_1 \oplus \cdots \oplus L_k} \cong F^{L_1} \wedge \cdots \wedge F^{L_k}$, we get a Thom class $u_{L_1} \cdots u_{L_k} \in E^{2k}(F^{L_1 \oplus \cdots \oplus L_k})$. Moreover, this is in the image of $E^{2k}(X^V)$, and so specifies a well-defined Thom class for $X^V$.

Spelled out in this detail, it might perhaps be easier to use the complex orientation to compute $E^*(X^V)$ and directly establish a Thom isomorphism, and thus a Thom class, since this is essentially the argument for the Leray-Hirsch theorem above.

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There was also a question in the comments about $k$ vs. $2k$. Just consider ordinary homology: $H^\text{odd}(\mathbb{C}P^\infty) = 0$, so we really do want $H^{2k}$. This is an artifact of the fact that there are real spheres of odd dimension that complex geometry cannot see.