Let be the Thom spectrum for oriented cobordism, so can be obtained as a homotopy colimit

where the th space is the Thom space of the (universal) vector bundle over . By the Thom-Pontryagin theorem, the homotopy groups are isomorphic (as a graded ring) to the cobordism ring of oriented manifolds. The goal of the next few posts is to discuss some of the classical results on the oriented cobordism ring. This post will handle the easiest case; since it is somewhat analogous to the situation for complex cobordism, it is a bit brief.

In the past, we described Milnor’s computation of (the complex cobordism ring), and showed that . The situation for is somewhat more complicated, as there will be torsion; as we will see, however, it is only 2-torsion that occurs, and not too wild of a sort. Let’s start by noting what is.

Theorem 1 (Thom)We have

where the can be taken to be the even-dimensional .

This result is a corollary of Serre’s work, which shows that rational stable homotopy is equivalent to rational homology. In other words, we have an isomorphism

(where the last isomorphism is the Thom isomorphism). Now is a polynomial ring on variables as above; this follows from the following computation:

Proposition 2Let be a ring containing . Then is a polynomial ring on generators . These generators are obtained as the image of generating elements in .

*Sketch proof:* If we grant the analogous theorem for (that is a polynomial ring on the generators of , without hypotheses on ), then we can deduce the result as follows. There is a “forgetful” map as well as a “complexification” map . The composite is homotopy equivalent to multiplication by on the H space . Consequently, we have a diagram in homology

where the composite is multiplication by two. It follows that is a retract of , because we have assumed .

The upshot of all this is that is a quotient of : it is the image under the realification map , or equivalently the image in homology of the operator sending a (stable) complex vector bundle to . In cohomology, the map

corresponding to the map annihilates the odd-dimensional Chern classes: if is any complex vector, then for odd, as one checks by writing as a direct sum of line bundles; by contrast, the even-dimensional Chern classes are pretty much arbitrary. For if , then

which shows that the even-dimensional Chern classes (as the symmetric functions of ) can be algebraically independent, while the odd-dimensional Chern classes are zero.

It follows that is a polynomial ring in the images of the even-dimensional Chern classes . The corresponding Hopf algebra is the subalgebra of generated by the even-dimensional classes in .

**1. Computing the homology**

Our next goal is to show that there is no odd torsion in the oriented cobordism ring . In order to do this, we will use the Adams spectral sequence (mod an odd prime). Recall that this is a spectral sequence

Our claim is that we can completely work out the term in the same way as we did for , by working out the coaction of the dual Steenrod algebra Recall that, for *odd*, we have the isomorphism

In fact, we have an analogous situation as with complex cobordism:

Proposition 3Let be the subHopfalgebra (i.e., not including the ). As a comodule over , we have

where is in degree .

In fact, let’s start by taking ; this has a basis . Then by the previous result, we have that, under the inclusions ,

As we saw earlier, the coaction of the dual Steenrod algebra in is via, for any ,

This reflects the fact that, when using Steenrod th powers, one can get from the cohomology class in degree to cohomology classes in degrees but nowhere else. Also,

if . The “dots” indicate tensor products where both have smaller degree (“decomposable elements”).

Now, we have an equivalence coming from the equivalence and the fact that the Thom space of is homotopy equivalent to it. In particular,

where the element in degree zero is the Thom class, corresponding to .

Let be the image of under this equivalence ; somewhat confusingly, under the Thom isomorphism with , it actually corresponds to . When computing the coaction of , the Thom isomorphism should be avoided, though, at least here. Then, we find that , where in the homology ring (as it corresponds to the Thom class). This follows by the Thom isomorphism and the analogous computation for . The coaction is thus given by a shift of the earlier formulas:

Motivated by this, we can define a map

which is defined via

here the first map is the coaction (observe that is a comodule over , as it has even degrees) and the second map sends for and for . This is a morphism of comodule algebras, and it is an isomorphism on indecomposables by the above formulas (1) and (2).

In general, it’s worth noting that comodule algebras tend to have a simple structure; we’ll see more of this in the next post.

**2. The Adams spectral sequence for MSO**

We can now continue more or less repeating the story for for (but only at an *odd* prime). Namely, we saw in the previous section that . The change-of-rings formula now gives, as before:

Here, in the exterior algebra , we have

As before, of an exterior algebra is a bigraded polynomial algebra: for bigraded via . Here can be represented in the cobar complex of via . In particular, we get from (3) and the Künneth formula that

where and .

Dimensional considerations now force it to degenerate. Observe that the vertical lines are chains of multiplication by , which corresponds to multiplication by in homotopy: it follows that the extension problems which arise are all of the same sort, and that the homotopy groups are devoid of -torsion. We conclude in fact:

Theorem 4.

In fact, we can argue as before: the indecomposable elements in in each degree form an abelian group which is either free of rank one over or zero. This itself forces the isomorphism, as taking the to be indecomposable generators produces a map

which is surjective as it hits indecomposables, and is injective because when tensored with both rings have the same graded dimension.

In the next post, we’ll begin analyzing the situation at the prime .

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