is a spectrum and, for the same reasons as , is in fact a commutative ring spectrum (even an ring spectrum).
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.
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.
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.
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.
unless , in which case one gets .
Proof: Induction on (when it is clear). Suppose . Then, we clearly have , so the inductive hypothesis implies that .
This means that is forced to be .
The are defined as the dual elements to (with respect to the Serre-Cartan basis of ).
Here the dots signify terms that do not include or .
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 .
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 .
I've just been writing reduced homology for emphasis: that's the natural thing for spectra, anyway.
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.
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 .
where the signify decomposable elements in .
When we analyze as a comodule, we will get something similar, but it will not be quite so simple.
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.
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 .
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.
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.
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.
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 .
This completes the proof of Thom's theorem.