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…)