The next goal in this series of posts is to understand fully the computation of the complex cobordism ring , which is due to Milnor. To start with, though, let’s consider a simpler example: the spectrum . The spectrum is obtained in the same way that is, but with the Thom spaces of the universal -dimensional *real* bundles over . In other words, if is the universal -dimensional bundle, we define

(where means Thom space) and define maps

coming from the map classifying . The homotopy colimit of the sequence

is a spectrum and, for the same reasons as , is in fact a commutative ring spectrum (even an ring spectrum).

One reason to care about is the following result:

Theorem 1 (Thom)The homotopy groups (a graded ring) is the cobordism ring of unoriented manifolds.

In other words, to describe , we can also use geometry: an element of is a compact -manifold modulo the relation of *cobordism*: two -manifolds are cobordant if there is an -manifold-with-boundary such that . From a geometric point of view, it is thus interesting to determine . It turns out that we can do so using homotopy-theoretic, algebraic methods; this was what Thom showed.

How might we compute the homotopy groups? In general, the homotopy groups of any space are extremely difficult to compute, but in this case we can get a complete answer for two reasons: first, it’s stable homotopy groups we’re after, not just homotopy groups; second, much more importantly, is a fairly simple spectrum.

In fact, we have:

Theorem 2 (Thom)is a wedge of Eilenberg-MacLane spectra (and its shifts) and is a polynomial ring on variables for all such that is not a power of .

This is the theorem I’d like to discuss today. It is definitely a much easier computation than that of , but it will be a toy example.

**1. The homology of : part 1**

The strategy for showing that splits as a sum of shifts of is to look at its *(co)homology*. It turns out that the cohomology of is a free module over the Steenrod algebra . In view of this, and the Hurewicz theorem, we’ll be able to see that itself must be a sum of ‘s.

But first we need to know what the homology of is, and we’ll need as a ring as well as a graded group. Fortunately, we have a fairly explicit homotopy presentation of as the homotopy colimit of , and we can work out the homology of each of these explicitly via the Thom isomorphism.

Let’s start by recalling that

where each in degree is dual to the th power of the generator . This handles . The homology of the spaces in general can be computed as follows:

Theorem 3The map establishes an equivalence

In other words, a basis for is given by the products where .

In the limit, we find that is a polynomial ring (the element becomes the unit).

By the Thom isomorphism, the homology of the spectrum is more or less going to be the homology of , so it seems we’ve answered the first question. But we need a bit more: we need the action of the dual Steenrod algebra , and this is in fact the main part. To understand the action of on , let’s start by working it out on . It’ll be easier to first do the dual computation.

Proposition 4Let be the generator. Let be a sequence of nonzero elements. Then

unless , in which case one gets .

*Proof:* Induction on (when it is clear). Suppose . Then, we clearly have , so the inductive hypothesis implies that .

The claim is thus that the only Steenrod square that doesn’t annihilate is . This follows from the fact that if is the total Steenrod square, then

This means that is forced to be .

With this in mind, we can write down what is as a *comodule* over . Let’s recall the structure of the dual Steenrod algebra , which was determined by Milnor. There are elements of degree , such that

The are defined as the dual elements to (with respect to the Serre-Cartan basis of ).

We have a comodule structure map

Another way of phrasing the previous result is that the coefficient of in is if , and zero otherwise. That is,

Here the dots signify terms that do not include or .

**2. The homology of , II**

Let’s now finish working out the homology of as a comodule over . We know that if is the Thom space of the universal bundle on , then (because the -bundle over is contractible), and that the Thom class in corresponds to the generator .

In other words, we have that

where each is in degree and corresponds under the Thom isomorphism to : equivalently, under the equivalence , corresponds to . In particular,

where each is in degree ; this is because of the desuspension in . Since is free on the appropriately ordered products of the , and we have a commutative diagram (where all homology is with -coefficients)

we find inductively that the vector space is free on the products with . Observe that the desuspensions made in mean that we don’t have to change degrees anywhere. Moreover, becomes the unit becomes acts as the unit in .

So, putting everything together, we find:

Proposition 5We have an isomorphism of rings

I’ve just been writing reduced homology for emphasis: that’s the natural thing for spectra, anyway.

The next goal is to figure out the action of on this homology. The map is just the map

and it is in particular a morphism of rings. So we just need to figure out where each goes.

In fact, comes from where each is in degree : the element corresponds to . By naturality, we might as well work out the coaction of on . Let’s work it out mod decomposables.

Proposition 6The coaction sends

The dots signify terms not involving or .

*Proof:* To see this, we just have to remember that is a desuspension of , and consequently is just a shift of . Consequently, the ‘s are corresponding to the ‘s under the identification . If we keep this in mind, then the formula is a consequence of the previous analysis for .

Now, if we recall that is just the unit element of under the map , then we find that the coaction sends

where the signify decomposable elements in .

The main result is:

Theorem 7As an -comodule and an algebra, there is an isomorphism

When we analyze as a comodule, we will get something similar, but it will not be quite so simple.

*Proof:* The strategy is to produce a map of comodule-algebras . We can do this very easily: it’s the same thing as a map of plain algebras

because the tensor product is a “coinduced” comodule and has a similar universal property (for mapping into it) that induced modules have in ordinary algebra. We define this map via if and zero otherwise.

The map of comodule-algebras is then obtained via the composition

and it is this composition we want to prove is an isomorphism. So far, we see that it’s a map between graded algebra of the same dimension in each degree, since .

To show that the composite map is an isomorphism, we’re going to have to use a commonly used technique: we’ll look at indecomposable elements. In dimension , the indecomposable elements in is spanned the image of . We know, moreover, from our earlier computations that modulo decomposables,

This is, in either case, indecomposable in degree in . In particular, the map sends indecomposables to indecomposables, so by induction on the degree we find that is a surjection. Since the domain and the codomain have the same dimension in each degree, must be an isomorphism.

**3. The homotopy of **

In the previous sections, we worked out completely the homology of as a comodule over the coalgebra , as well as an algebra in its own right. We know want to work out the spectrum , and prove the theorems stated at the start of the blog post.

So, we want to show that is a wedge of Eilenberg-MacLane spectra.

Proposition 8Choose a basis of the graded piece . Then we have a decomposition of spectraSuppose is a connective spectrum, whose homotopy groups are all 2-torsion. Suppose is a co-induced module over , i.e. for a graded -vector space whose graded pieces are finite-dimensional.

*Proof:* The strategy will be to produce a map which is an isomorphism on -homology, and which will then be forced to be an equivalence.

In fact, we find that is free as a module over the Steenrod algebra, . The dual basis elements are cohomology classes in dimension which define a map

This map is an isomorphism on cohomology by construction: the cohomology of an Eilenberg-MacLane spectrum is free on .

Dualizing, we find that it is an isomorphism on -homology. It follows that the cofiber, which is a connective spectrum with 2-torsion homotopy groups, has trivial -homology. This means that the cofiber is contractible: for, if its first nonzero homotopy group were in degree , then we would have .

So, in particular, applying this to , we find that is a wedge of shifts of . This means in particular that the Hurewicz map

is an injection, and that its image must consist of the primitive elements: these are given by

This completes the proof of Thom’s theorem.

May 23, 2012 at 6:13 pm

[…] of this series of posts is to understand the computation of the complex cobordism ring . In the computation of the unoriented cobordism ring , we did this by working out as a comodule over the dual Steenrod algebra , and observing that it […]

May 24, 2012 at 4:05 pm

[…] proof of this will be analogous to the computation of . In fact, the idea is essentially that, by the Thom isomorphism […]