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

$\begin{useimager} \[ \begin{tikzcd} E \wedge \Th(\xi) \ar[r, "E \wedge \Delta"] & E \wedge \Th(\xi) \wedge X_+ \ar[r, "E \wedge u \wedge X_+"] & E \wedge E \wedge X_+ \ar[r, "\mu \wedge X_+"] & E \wedge X_+ \end{tikzcd} \] \end{useimager}$

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

$\begin{useimager} \[ \begin{tikzcd} F \wedge A \ar[r, "F \wedge \iota \wedge A"] & F \wedge E \wedge A \ar[r, "F \wedge \varphi"] & F \wedge E \wedge B \ar[r, "F \wedge f \wedge B"] & F \wedge F \wedge B \ar[r, "\mu_F \wedge B"] & F \wedge B. \end{tikzcd} \] \end{useimager}$

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

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

$\begin{useimager} \[ \begin{tikzcd} MU \wedge \Th(\xi) \ar[r, "MU \wedge \Delta"] & MU \wedge \Th(\xi) \wedge X_+ \ar[r, "MU \wedge u \wedge X_+"] & MU \wedge MU \wedge X_+ \ar[r, "\mu \wedge X_+"] & MU \wedge X_+ \end{tikzcd} \] \end{useimager}$

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

$\begin{useimager} \[ \begin{tikzcd} \gamma \oplus \xi \ar[d] \ar[r] & \gamma \oplus \xi \oplus 0 \ar[d] \ar[r] & \gamma \oplus \gamma \oplus 0 \ar[d] \ar[r] & \gamma \oplus 0 \ar[d]\\ BU \times X \ar[r, "BU \times \Delta"] & BU \times X \times X \ar[r, "BU \times f \times X"] & BU \times BU \times X \ar[r, "\oplus \times X"] & BU \times X \end{tikzcd} \] \end{useimager}$

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

ring maps $MU \to E$ ; and
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

complex orientations of $E$ ; and
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.