This is the fourth in a series of posts on the oriented cobordism ring. In the previous posts, we showed that the oriented cobordism ring had no odd torsion, and determined the structure of the mod 2 cohomology. The purpose of this post is to tie this results together by describing $MSO_{(2)}$ and then applying that to compute a few low-dimensional cobordism groups.

I stated earlier that we would prove the following result using the Adams spectral sequence:

Theorem 12 (Wall) All the torsion in the oriented cobordism ring ${\Omega_{SO} \simeq \pi_* MSO}$ has order two: that is, each cobordism group ${\Omega_{SO}^k}$ is a direct sum of copies of ${\mathbb{Z}}$ and ${\mathbb{Z}/2}$.

In fact, once we know that there is no odd torsion (which we proved using the ASS), we don’t need the Adams spectral sequence to prove this. We can prove directly the following result, which will imply Wall’s theorem: the spectrum ${MSO_{(2)}}$ is equivalent to a wedge of shifts of ${H \mathbb{Z}_{(2)}}$ and ${H \mathbb{Z}/2}$. To see this, we use the main result of the previous post: the mod 2 cohomology of ${MSO_{(2)}}$ (or ${MSO}$, same thing) is a direct sum of copies of ${\mathcal{A}_2}$ and ${\mathcal{A}_2/\mathcal{A}_2 \mathrm{Sq}^1}$.

Now, to show that ${MSO_{(2)}}$ is as claimed, it suffices to show that there is a map $\displaystyle MSO_{(2)} \rightarrow \bigoplus_{i \in F} H \mathbb{Z}_{(2)}[d_i] \oplus \bigoplus_{j \in G} H \mathbb{Z}/2[d_j],$

inducing an isomorphism in mod 2 homology: in fact, one such is necessarily an equivalence in the since everything is localized at 2 (at least for connective spectra).

To produce such a map, we start by choosing generators ${\alpha_j}$ for ${H^*(MSO; \mathbb{Z}/2)}$: we can choose the generators ${\alpha_j}$ such that they are of two types: for ${j \in J_1}$, ${\alpha_j}$ is acted upon freely by ${\mathcal{A}_2}$; for ${j \in J_2}$, ${\alpha_j}$ generates a submodule isomorphic to ${\mathcal{A}_2/\mathcal{A}_2 \mathrm{Sq}^1}$. In other words, the ${\alpha_j}$ together define an isomorphism of ${H^*(MSO; \mathbb{Z}/2)}$ with a direct sum of copies of ${\mathcal{A}_2 }$ and ${\mathcal{A}_2/\mathcal{A}_2 \mathrm{Sq}^1}$.

Altogether, we get a map of spectra $\displaystyle MSO_{(2)} \rightarrow \bigoplus_j H \mathbb{Z}/2[d_j]$

by the universal property of an Eilenberg-MacLane spectrum. This is not quite what we want: we want a bunch of ${H \mathbb{Z}}$‘s as well. For ${j \in J_2}$, ${\alpha_j}$ is annihilated by the Bockstein coboundary, so we can lift the ${j}$th map, ${j \in J_2}$, $\displaystyle MSO_{(2)} \rightarrow H \mathbb{Z}/2[d_j]$

to a map $\displaystyle MSO_{(2)} \rightarrow H \mathbb{Z}_{(2)}[d_j].$

Together, these define a map $\displaystyle \Phi: MSO_{(2)} \rightarrow \bigoplus_{j \in J_1} H \mathbb{Z}/2[d_j] \oplus \bigoplus_{j \in J_2} H \mathbb{Z}_{(2)}[d_j].$

Actually, a priori we would be mapping to the direct product, but it is the same as the direct sum since the cohomology is finite in each degree. Since the cohomology of ${H\mathbb{Z}_{(2)}}$ is ${\mathcal{A}_2/\mathcal{A}_2 \mathrm{Sq}^1}$, it follows that ${\Phi}$ is an equivalence on mod 2 cohomology, and thus an equivalence.

We have now proved a result we have been stating for a while:

Theorem 13 (Wall) The spectrum ${MSO_{(2)}}$ is equivalent to a direct sum of shifts of copies of ${H \mathbb{Z}_{(2)}}$ and ${H \mathbb{Z}/2}$.

In particular, since there is no odd torsion, the homotopy groups of $MSO$ (i.e., the cobordism groups) are direct sums of ${\mathbb{Z}}$‘s and ${\mathbb{Z}/2}$‘s. This in particular means that the cobordism ring is actually commutative (rather than graded-commutative)

Computations

The neat thing about Wall’s theorem is that, while the decomposition it gives is somewhat inexplicit, it’s not actually hard to work out what the decomposition is in low dimensions. In particular, we can work out some low-dimensional cobordism groups.

The idea is that we do know where the summands of ${H \mathbb{Z}}$ (or ${H \mathbb{Z}_{(2)}}$) come into the decomposition of ${MSO}$ at ${2}$: namely, there is one summand of ${H \mathbb{Z}[4d]}$ for each partition of ${d}$. This follows from the determination of the rational cobordism ring ${\pi_* MSO \otimes \mathbb{Q} \simeq \mathbb{Q}[x_4, x_8, \dots ]}$. To work out where the ${H \mathbb{Z}/2}$‘s live, we can inductively look at the dimensions in each grading of $\displaystyle H^*(MSO; \mathbb{Z}/2) \simeq H^*(BSO; \mathbb{Z}/2) \simeq \mathbb{Z}/2[w_2, w_3, w_4, \dots, ].$

Let’s draw a table indicating elements in ${H^*(MSO; \mathbb{Z}/2)}$; these will be multiples of the Thom class ${t}$ by an element of ${H^*(BSO; \mathbb{Z}/2)}$. We compare it with a table of elements in ${H^*(H \mathbb{Z}; \mathbb{Z}/2)}$, which is generated by the fundamental class ${\iota_0}$. It has a (Serre-Cartan) basis of monomials $\displaystyle \mathrm{Sq}^{i_1} \dots \mathrm{Sq}^{i_n} \iota_0, \quad i_k \geq 2i_{k-1}, \ i_n \neq 1.$ So far, in degrees up to three, all the cohomology is accounted for by ${H \mathbb{Z}}$. In particular, ${\pi_i MSO_{(2)} = \pi_i H \mathbb{Z}_{(2)}}$ for ${i = 0, 1, 2,3}$. Since there is no odd torsion in the cobordism ring by Milnor’s theorem, we have computed the first three cobordism groups, $\displaystyle \Omega_{SO}^0 = \mathbb{Z}, \ \Omega_{SO}^1 = 0 , \ \Omega_{SO}^2 = 0 , \ \Omega_{SO}^3 = 0.$

Of these, the zeroth through second cobordism groups are elementary to evaluate (using e.g., the classification of surfaces). It is nontrivial that the third oriented cobordism group vanishes; this is a theorem of Rohlin which also follows from the present machinery.

Anyway, we can continue. We see in dimension four that ${H \mathbb{Z}}$ in dimension zero doesn’t account for all the cohomology: this is reasonable, because we know there has to be another ${H \mathbb{Z}}$ summand. So let’s draw the table of cohomologies again, with ${\iota_4}$ the fundamental cohomology class of ${H \mathbb{Z}}$: Note that there is a discrepancy in dimension 5 again: the cohomology of ${H \mathbb{Z} \oplus H \mathbb{Z}}$ fails to account for the cohomology of ${MSO}$ in this degree. So we need to add a ${H \mathbb{Z}/2}$, whose cohomology is free on an element ${\iota_5}$ in degree five. In particular, $\displaystyle \Omega_{SO}^4 = \mathbb{Z} \oplus \mathbb{Z}, \quad \Omega_{SO}^5 = \mathbb{Z}/2.$

We can continue the table. (We will stop drawing the earlier terms.) It follows that the dimensions agree in these two dimensions, so we don’t need to add any more ${H \mathbb{Z}/2}$‘s. We find: $\displaystyle \Omega_{SO}^6 = \Omega_{SO}^7 = 0.$

This can evidently be continued. The fact that ${\Omega_{SO}^7 =0}$, incidentally, was used in Milnor’s mod ${7}$ invariant on (appropriate) 7-manifolds that he used to distinguish certain manifolds as exotic spheres; see this post.