(This is the second in a series of posts intended for me to try to understand the connection between stable homotopy theory and formal group laws.)

Last time, we introduced the notion of a complex-oriented cohomology theory and saw that we could imitate the classical theory of Chern classes in one such. In this post, I’d like to describe the universal example of a complex-oriented cohomology theory: complex cobordism. This is going to play a very special role in the next few posts.

To unravel this, let’s try to recall what a complex orientation was. It was a choice of Thom classes of complex vector bundles, functorial in the bundle and multiplicative. For starters, let’s focus now on the functoriality. Let ${E}$ be a cohomology theory represented by a spectrum ${E}$. Then since there is a universal ${n}$-dimensional vector bundle ${\zeta_n \rightarrow BU(n)}$, it follows that a functorial choice of Thom classes for ${n}$-dimensional vector bundles is the same as a Thom class for ${\zeta_n}$. So, all we need to give is an element of ${\widetilde{E}^*(T(\zeta_n))}$. If we (and we henceforth do this) normalize things such that the Thom class of the ${n}$-th degree element is in degree ${n}$, then we have to give an element of ${\widetilde{E}^n(T(\zeta_n))}$.

Definition 1 The spectrum ${MU(n)}$ is ${\Sigma^{-2n}T(\zeta_n)}$.

So another way of saying this is that we should have a map of spectra

$\displaystyle MU(n) \rightarrow E.$

There is a map ${S^{2n} \rightarrow T(\zeta_n)}$ which comes from fixing a basepoint in ${BU(n)}$. So, in other words, to give a functorial complex orientation for ${n}$-dimensional complex vector bundles is to give an element of ${\widetilde{E}^n(T(\zeta_n))}$ which restricts to the generator of ${\widetilde{E}^n(S^{2n})}$. (To check that an element is a Thom class, we only need restrict it to one fiber in each connected component of the base.)

1. ${MU}$

Let’s try to say a little more about these spectra ${MU(n)}$.

The spaces ${MU(n)}$ come with additional structure. For ${m, n}$, there is a map

$\displaystyle MU(n) \wedge MU(m) \rightarrow MU(n + m)$

which comes from the map

$\displaystyle BU(n) \times BU(m) \rightarrow BU(n+m)$

classifying the direct sum of vector bundles. Moreover, for each ${n}$, there is a map

$\displaystyle MU(n) \rightarrow MU(n+1)$

which comes from the map

$\displaystyle BU(n) \rightarrow BU(n+1)$

which corresponds functorially to adding a trivial summand to a vector bundle. In other words, it classifies ${\zeta_n \oplus \mathbb{C}}$, and ${T(\zeta_n \oplus \mathbb{C}) = T(\zeta_n) \wedge S^2}$. So this is just a desuspension of the map ${T(\zeta_n) \wedge S^2 \rightarrow T(\zeta_{n+1} )}$.

Morally, the conclusion should be now that the spaces ${MU(n)}$ fit into a ring spectrum representing a multiplicative cohomology theory. This isn’t quite obvious yet. What is definitely true is that we can define a spectrum ${MU}$ from the maps ${ MU(n) \rightarrow MU(n+1)}$: in other words, we have to take the homotopy colimit of the sequence

$\displaystyle MU(0) \rightarrow MU(1) \rightarrow MU(2) \rightarrow \dots .$

If you like to think of a spectrum as a sequence of pointed spaces ${X_n}$ with maps ${S^1 \wedge X_n \rightarrow X_{n+1}}$, then you could take ${X_{2n} = T(\zeta_n), X_{2n+1} = T(\zeta_n) \wedge S^1}$.

Anyway, it turns out that ${MU}$ actually does fit into a ring spectrum. I don’t know if there’s a good way to see this formally using the homotopy colimit presentation (perhaps someone can enlighten me?); however, if one uses the model of symmetric spectra, then one can explicitly see that ${MU}$ is a symmetric ring spectrum. So we get more: not only is ${MU}$ a ring spectrum (an algebra object in the stable homotopy category), but it’s an ${E_\infty}$ ring spectrum (a ring spectrum with a coherently commutative and associative multiplication law). See for instance Schwede’s untitled book on the subject.

This spectrum is called complex cobordism.

2. Complex orientations again

Let’s return to the original topic: we were trying to rewrite the definition of a complex orientation of a ring spectrum ${E}$ (corresponding to a multiplicative cohomology theory). We saw that a complex orientation on ${E}$ required maps

$\displaystyle MU(n) \rightarrow E$

for each ${n}$. In other words, elements ${\theta_n \in \widetilde{E}^*(MU(n))}$. These would have to be compatible in that the element ${\theta_n \theta_m}$ is the pull-back of ${\theta_{n+m}}$ in ${\widetilde{E}^*(MU(n+m))}$. Moreover, to be actual Thom classes, they’d have to restrict to the unit map on ${MU(0) = S^0}$.

So to give a complex orientation on ${E}$ is to give maps of spectra ${\theta_n: MU(n) \rightarrow E}$ for each ${n}$ such that the following diagram is always homotopy commutative

and also the restriction of ${\theta_{n+1}: MU(n+1) \rightarrow E}$ to ${MU(n)}$ is ${\theta_n}$.

These will actually fit into a morphism of ring spectra ${MU \rightarrow E}$. This isn’t totally obvious, so let’s give an argument. A complex orientation gives us a compatible family of maps ${MU(n) \rightarrow E}$, or compatible elements in ${\widetilde{E}^*(MU(n))}$. Using the Thom isomorphism, we find that ${\widetilde{E}^*(MU(n)) \simeq \widetilde{E}^*(BU(n)) = (\pi_* E ) [[c_1, \dots, c_n]]}$ and the transition maps are actually surjections. Consequently, by the Milnor exact sequence

$\displaystyle \widetilde{E}^*(MU) = \varprojlim \widetilde{E}^*(MU(n)).$

In other words, the ${\lim^1}$-terms are zero, and a map ${MU \rightarrow E}$ in the homotopy category is determined by its restriction to the ${MU(n)}$. So, given a complex orientation, we get an honest map of spectra

$\displaystyle MU \rightarrow E$

which is unique in restricting to ${\theta_n}$ on ${MU(n)}$.

The thing to see now is that the map ${MU \rightarrow E}$ is an actual map of ring spectra. In other words, that there is a commutative diagram

This follows because from the two directions of the diagram, one gets two elements of ${\widetilde{E}^*(MU \wedge MU)}$ which restrict to the same element of ${\widetilde{E}^*(MU(n) \wedge MU(m))}$ for each ${n, m}$. Using the Milnor exact sequence, again, we find that this means the diagram commutes.

From this, we get:

Theorem 2 If ${E}$ is a ring spectrum, there is a bijection between complex orientations on ${E}$ and maps of ring spectra ${MU \rightarrow E}$.

In particular, ${MU}$ itself is complex-oriented, from the identity map ${MU \rightarrow MU}$. What does this correspond to? This is the “tautological” orientation on complex cobordism. Given a complex vector bundle ${V \rightarrow X}$, classified by a map ${X \rightarrow BU(n)}$, then the Thom class in ${\widetilde{MU}^*(T(V))}$ comes from the map

$\displaystyle T(V) \rightarrow MU(n)$

induced on Thom spaces from ${X \rightarrow BU(n)}$.

In any event, we saw in the last post that a complex orientation on a cohomology theory leads to a formal group law. What is the formal group law for complex cobordism? The remarkable answer, due to Quillen, is that it is the universal formal group law — in precisely the same way as $MU$ is universal among complex-oriented cohomology theories.

I’d like to try to describe the proof of Quillen’s theorem as completely as possible in the next few posts. As far as I know, there is no non-computational proof of this result, and two very computational (non-formal) inputs have to go into it: on the one hand, the structure of the underlying complex cobordism ring $\pi_* MU$ (which was worked out by Milnor using an Adams spectral sequence computation); on the other hand, the structure of the universal one-dimensional commutative formal group law (which was worked out by Lazard). Then, we’ll need to figure out exactly how to compare the formal group laws associated to different complex-oriented theories.