(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 be a cohomology theory represented by a spectrum . Then since there is a universal -dimensional vector bundle , it follows that a functorial choice of Thom classes for -dimensional vector bundles is the same as a Thom class for . So, all we need to give is an element of . If we (and we henceforth do this) normalize things such that the Thom class of the -th degree element is in degree , then we have to give an element of .

Definition 1The spectrum is .

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

There is a map which comes from fixing a basepoint in . So, in other words, to give a functorial complex orientation for -dimensional complex vector bundles is to give an element of which restricts to the generator of . (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. **

Let’s try to say a little more about these spectra .

The spaces come with additional structure. For , there is a map

which comes from the map

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

which comes from the map

which corresponds functorially to adding a trivial summand to a vector bundle. In other words, it classifies , and . So this is just a desuspension of the map .

Morally, the conclusion should be now that the spaces 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 from the maps : in other words, we have to take the homotopy colimit of the sequence

If you like to think of a spectrum as a sequence of pointed spaces with maps , then you could take .

Anyway, it turns out that 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 is a symmetric ring spectrum. So we get more: not only is a ring spectrum (an algebra object in the stable homotopy category), but it’s an 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 (corresponding to a multiplicative cohomology theory). We saw that a complex orientation on required maps

for each . In other words, elements . These would have to be compatible in that the element is the pull-back of in . Moreover, to be actual Thom classes, they’d have to restrict to the unit map on .

So to give a complex orientation on is to give maps of spectra for each such that the following diagram is always homotopy commutative

and also the restriction of to is .

These will actually fit into a morphism of *ring spectra* . This isn’t totally obvious, so let’s give an argument. A complex orientation gives us a compatible family of maps , or compatible elements in . Using the Thom isomorphism, we find that and the transition maps are actually surjections. Consequently, by the Milnor exact sequence

In other words, the -terms are zero, and a map in the homotopy category is determined by its restriction to the . So, given a complex orientation, we get an honest map of *spectra*

which is unique in restricting to on .

The thing to see now is that the map 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 which restrict to the same element of for each . Using the Milnor exact sequence, again, we find that this means the diagram commutes.

From this, we get:

Theorem 2If is a ring spectrum, there is a bijection between complex orientations on and maps of ring spectra .

In particular, itself is complex-oriented, from the identity map . What does this correspond to? This is the “tautological” orientation on complex cobordism. Given a complex vector bundle , classified by a map , then the Thom class in comes from the map

induced on Thom spaces from .

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 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 (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.

## Leave a Reply