The next goal of this series of posts (started here) is to analyze the oriented cobordism spectrum {MSO} at the prime 2; the main result is that there is a splitting of {MSO_{(2)}} into a direct sum of copies of {H\mathbb{Z}_{(2)}} (the torsion-free part) and {H \mathbb{Z}/2} (the torsion-part). In particular, it will follow that there is only torsion of order two in the cobordism ring — since we showed last time that there was no odd torsion. We will see this using the Adams spectral sequence at the prime {2}, once we’ve figured out what {H_*(MSO; \mathbb{Z}/2)} looks like as a comodule over the dual Steenrod algebra. This, however, is apparently somewhat tricky to do directly.

In order to get there, we’ll need a bit of algebraic machinery (which we state in a dual context). Recall that a graded vector space {V} is called connected if {V_0} is one-dimensional and {V_i = 0} for {i < 0}. The next result provides a sufficient criterion for a module over a graded, connected Hopf algebra to be free.

Theorem 5 (Milnor-Moore) Let {A} be a connected, graded Hopf algebra over a field {k}, and let {M} be a graded, connected {{A}}-module which is simultaneously a coalgebra (in such a way that {M \rightarrow M \otimes_k M} is an {A}-homomorphism). Let {1 \in M_0} be a generator, and suppose the map of {A}-modules

\displaystyle A \rightarrow M, \quad a \mapsto a . 1

is a monomorphism. Then {M} is a free graded {A}-module.

This is a pretty surprising result, as a relatively minor hypothesis (coalgebra, and the action on {1} is free) leads to freeness of the whole thing. The idea of the proof is going to be to produce generators of {M} by lifting a vector space basis of {\overline{M} = M \otimes_A k}. The fact that these generators are forced to be linearly independent is an unexpected consequence of the coalgebra structure; the graded connectedness will be used to make certain inductive arguments. (more…)