Complex oriented cohomology theoriesOrientations

3 Orientations

Definition 3.1

Let $E$ be a ring spectrum, and $\xi : V \to X$ a vector bundle. Then $V$ is $E$-oriented if there is a “Thom class” $u: \mathrm{Th}(\xi ) \to E$ such that the induced map of $E$-modules

is an isomorphism.

This induces isomorphisms

$E_*(X_+) \cong E^*(\mathrm{Th}(\xi )),\quad E^*(X_+) \cong E^*(\mathrm{Th}(\xi ))$

induced by cupping and capping with $u$ (using the Thom diagonal $\Delta$).

Lemma 3.2

Let $f: E \to F$ be a map of ring spectra. Then an $E$-orientation of $\xi$ gives rise to an $F$-orientation of $\xi$ by composition.

Proof
[Proof sketch] The key input here is that an $E$-module map $\varphi : E \wedge A \to E \wedge B$ functorially induces an $F$-module map $F \wedge A \to F \wedge B$ via the composite

Crucially, this construction does not make use of any coherence; if we had some sort of coherence, we could simply apply $F \wedge _E (-)$. One checks that the Thom isomorphism for $F$ is induced from that for $E$ via this procedure. Functoriality then ensures the resulting map is also an equivalence.

Proof

Definition 3.3

A cohomology theory $E$ is complex oriented if there is a choice of a Thom class $u_\xi$ for every complex vector bundle $\xi : V \to X$ that

1. is functorial under pullbacks, i.e. $f^* u_\xi = u_{f^* \xi }$; and

2. sends direct sums to products, i.e. $u_{\xi \boxplus \zeta } = u_\xi \smile u_\zeta$.

Example 3.4

$MU$ is complex oriented. Indeed, if $\xi : V \to X$ is a complex vector bundle, it is classified by a map $f: X \to BU$, and $\xi$ is the pullback of the universal bundle. Applying $\mathrm{Th}$ gives a map

$u_\xi = \mathrm{Th}(f): \mathrm{Th}(\xi ) \to MU.$

We claim this is a Thom class. We have to show that the composite

is an isomorphism. This map is obtained by applying $\mathrm{Th}$ to

Here $\gamma$ is the tautological bundle, and when we write $a \oplus b \oplus c$, we really mean $\pi _1^* a \oplus \pi _2^* b \oplus \pi _3^* c$.

We can describe the bottom map as sending $(v, x)$ to $(v + f(x), x)$. This has an inverse given by $(v, x) \mapsto (v - f(x), x)$, so is an isomorphism. Hence the induced map on Thom spaces is also an isomorphism.

The requirement that $u_{\xi \boxplus \zeta } = u_\xi \smile u_\zeta$ comes from the fact that the multiplication map $MU \times MU \to MU$ is induced by $\oplus : BU \times BU \to BU$.

Lemma 3.5

Let $E$ be a ring spectrum. Then there is a bijection between

1. ring maps $MU \to E$; and

2. complex orientations of $E$.

Proof
Since $MU$ is complex oriented, a ring map $MU \to E$ gives a complex orientation of $E$. Conversely, if $E$ is complex oriented, then since $MU$ is the Thom spectrum of a complex vector bundle, we get a Thom class $u: MU \to E$. This is a ring map since the product $MU \wedge MU \to MU$ is induced by the direct sum, and the requirement that the Thom class sends direct sums to products is exactly the statement that $MU \to E$ is a ring map.
Proof

Lemma 3.6

Let $E$ be complex oriented. Then there is a canonical isomorphism $E^*(\mathbb {CP}^n_+) \cong E^*[u]/u^{n + 1}$ and $E^*(\mathbb {CP}^\infty _+) \cong E^*[\! [u]\! ]$. Note that the choice of $u$ depends on the complex orientation of $E$.

Proof
We first construct the class $u$. For $n > 0$, we know that $\mathbb {CP}^n$ is the Thom space of the tautological bundle over $\mathbb {CP}^{n - 1}$. So there is a preferred class $u \in E^2(\mathbb {CP}^n)$. For $n = 1$, the Thom isomorphism theorem tells us this generates $E^*(\mathbb {CP}^1) \cong E^{* - 2}(S^0)$. This proves the lemma for $n = 1$. Note that inductively applying the theorem proves that $E^*(\mathbb {CP}^n_+) \cong E^*[u]/u^{n + 1}$ as groups, but we want the multiplicative structure as well.

For $n > 1$, consider the Atiyah–Hirzebruch spectral sequence for $E^*(\mathbb {CP}^n)$. We claim that $u$ is represented by a generator of $H^2(\mathbb {CP}^n; E^0) \cong E^0$. Indeed, if it weren't, then its image when pulled back along $\mathbb {CP}^1 \hookrightarrow \mathbb {CP}^n$ would also not be a generator, which contradicts our previous observation.

Now the $E^2$ page of the Atiyah–Hirzebruch spectral sequence for $E^*(\mathbb {CP}^n_+)$ is generated as a ring by $u$ and $E^*$, both of which are permanent. So the Atiyah–Hirzebruch spectral sequence degenerates and gives the desired isomorphism.

The $n = \infty$ case follows from taking the limit.

Proof

Theorem 3.7

There is a bijection between

1. complex orientations of $E$; and

2. classes $u \in E^2(\mathbb {CP}^\infty )$ that restrict to an $E^*$-module generator of $E^2(\mathbb {CP}^1) \cong E^0$.

Proof
[Proof sketch] Our previous computation showed that a complex orientation of $E$ induces such a class. The other direction is more roundabout. As in our previous argument, the class $u$ forces the Atiyah–Hirzebruch spectral sequence for $E^*(\mathbb {CP}^\infty _+)$ to degenerate at $E_2$. By the Kronecker pairing, the AHSS for $E_*(\mathbb {CP}^\infty _+)$ also has to degenerate at $E_2$.

Now since $H_*(MU)$ is generated as a ring by the image of $H_*(\Sigma ^{-2} MU(1)) = H_*(\Sigma ^{-2}\mathbb {CP}^\infty )$, we know the AHSS for $E_*(MU)$ also has to degenerate at $E_2$, and so does that for $E^*(MU)$. This gives us a preferred element in $E^0(MU)$ which is equivalently a map $MU \to E$ as desired. Playing the same game with $E^*(MU \wedge MU)$ shows that this map is a homotopy ring map.

Proof

Corollary 3.8

A even homotopy ring spectrum is always complex orientable.

Proof
The Atiyah–Hirzebruch spectral sequence for $\mathbb {CP}^\infty$ degenerates at $E_2$ for degree reasons.
Proof

Example 3.9

Complex $K$-theory is complex orientable.