The next goal in this series of posts is to understand fully the computation of the complex cobordism ring ${\pi_* MU}$, which is due to Milnor. To start with, though, let’s consider a simpler example: the spectrum ${MO}$. The spectrum ${MO}$ is obtained in the same way that ${MU}$ is, but with the Thom spaces of the universal ${n}$-dimensional real bundles over ${BO(n)}$. In other words, if ${\xi_n \rightarrow BO(n)}$ is the universal ${n}$-dimensional bundle, we define

$\displaystyle MO(n) = \Sigma^{-n} T(\xi_n),$

(where $T$ means Thom space) and define maps

$\displaystyle MO(n) \rightarrow MO(n+1)$

coming from the map ${BO(n) \rightarrow BO(n+1)}$ classifying ${\xi_n \oplus \mathbb{R}}$. The homotopy colimit of the sequence

$\displaystyle MO(0) \rightarrow MO(1) \rightarrow \dots$

is a spectrum and, for the same reasons as ${MU}$, is in fact a commutative ring spectrum (even an ${E_\infty}$ ring spectrum).

One reason to care about ${MO}$ is the following result:

Theorem 1 (Thom) The homotopy groups ${\pi_* MO}$ (a graded ring) is the cobordism ring of unoriented manifolds.

In other words, to describe ${\pi_* MO}$, we can also use geometry: an element of ${\pi_n MO}$ is a compact ${n}$-manifold ${M}$ modulo the relation of cobordism: two ${n}$-manifolds ${M, M'}$ are cobordant if there is an ${(n+1)}$-manifold-with-boundary ${W}$ such that ${\partial W = M \sqcup M'}$. From a geometric point of view, it is thus interesting to determine ${\pi_* MO}$. 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, ${MO}$ is a fairly simple spectrum.

In fact, we have:

Theorem 2 (Thom) ${MO}$ is a wedge of Eilenberg-MacLane spectra ${H\mathbb{Z}/2}$ (and its shifts) and ${\pi_* MO = \mathbb{Z}/2[x_2, x_4, \dots]}$ is a polynomial ring on variables ${x_i}$ for all ${i}$ such that ${i+1}$ is not a power of ${2}$.

This is the theorem I’d like to discuss today. It is definitely a much easier computation than that of ${\pi_* MU}$, but it will be a toy example.

1. The homology of ${MO}$: part 1

The strategy for showing that ${MO}$ splits as a sum of shifts of ${H\mathbb{Z}/2}$ is to look at its (co)homology. It turns out that the cohomology of ${MO}$ is a free module over the Steenrod algebra ${\mathcal{A}_2}$. In view of this, and the Hurewicz theorem, we’ll be able to see that ${MO}$ itself must be a sum of ${H \mathbb{Z}/2}$‘s.

But first we need to know what the homology of ${MO}$ is, and we’ll need as a ring as well as a graded group. Fortunately, we have a fairly explicit homotopy presentation of ${MO}$ as the homotopy colimit of ${MO(0) \rightarrow MO(1) \rightarrow \dots}$, and we can work out the homology of each of these explicitly via the Thom isomorphism.

Let’s start by recalling that

$\displaystyle H_* (\mathbb{RP}^\infty; \mathbb{Z}/2) \simeq \mathbb{Z}/2\left\{b_0, b_1, b_2, \dots\right\}$

where each ${b_i}$ in degree ${i}$ is dual to the ${i}$th power of the generator ${x \in H^1(\mathbb{RP}^\infty, \mathbb{Z}/2)}$. This handles ${BO(1) = \mathbb{RP}^\infty}$. The homology of the spaces ${BO(n)}$ in general can be computed as follows:

Theorem 3 The map ${(\mathbb{RP}^\infty)^n \rightarrow BO(n)}$ establishes an equivalence

$\displaystyle H_*(\mathbb{RP}^\infty; \mathbb{Z}/2)^{\otimes n}_{\Sigma_n} \simeq H_*(BO(n); \mathbb{Z}/2) .$

In other words, a basis for ${H_* (BO(n); \mathbb{Z}/2)}$ is given by the products ${b_{i_1} \dots b_{i_n}}$ where ${i_1 \leq \dots \leq i_n}$.

In the limit, we find that ${H_*(BO; \mathbb{Z}/2)}$ is a polynomial ring ${\mathbb{Z}/2[b_1, b_2, \dots, ]}$ (the element ${b_0}$ becomes the unit).

By the Thom isomorphism, the homology of the spectrum ${MO}$ is more or less going to be the homology of ${BO}$, so it seems we’ve answered the first question. But we need a bit more: we need the action of the dual Steenrod algebra ${\mathcal{A}_2^{\vee}}$, and this is in fact the main part. To understand the action of ${\mathcal{A}_2^{\vee}}$ on ${H_*(MO)}$, let’s start by working it out on ${\mathbb{RP}^\infty}$. It’ll be easier to first do the dual computation.

Proposition 4 Let ${x \in H^1(\mathbb{RP}^\infty, \mathbb{Z}/2)}$ be the generator. Let ${i_1, \dots, i_k}$ be a sequence of nonzero elements. Then

$\displaystyle \mathrm{Sq}^{i_k} \dots \mathrm{Sq}^{i_1} x = 0$

unless ${i_1 = 1, i_2 = 2, \dots, i_k = 2^{k-1}}$, in which case one gets ${x^{2^{k}}}$.

Proof: Induction on ${k}$ (when ${k =1}$ it is clear). Suppose ${\mathrm{Sq}^{i_k} \dots \mathrm{Sq}^{i_1} x \neq 0}$. Then, we clearly have ${\mathrm{Sq}^{i_{k-1}} \dots \mathrm{Sq}^{i_1} x \neq 0}$, so the inductive hypothesis implies that ${i_1 = 1, i_2 = 2, \dots, i_{k-1} = 2^{k-1}}$.

The claim is thus that the only Steenrod square that doesn’t annihilate ${x}$ is ${\mathrm{Sq}^{2^{k-1}}}$. This follows from the fact that if ${\mathrm{Sq}}$ is the total Steenrod square, then

$\displaystyle \mathrm{Sq}(x^{2^{k-1}}) = (\mathrm{Sq} x)^{2^{k-1}} = (x + x^2)^{2^{k-1}} = x^{2^{k-1}} + x^{2^k}.$

This means that ${i_k}$ is forced to be ${2^k}$. $\Box$

With this in mind, we can write down what ${H_* ( \mathbb{RP}^\infty)}$ is as a comodule over ${\mathcal{A}_2^{\vee}}$. Let’s recall the structure of the dual Steenrod algebra ${\mathcal{A}_2^{\vee}}$, which was determined by Milnor. There are elements ${\xi_i \in \mathcal{A}^{\vee}_{2}}$ of degree ${2^{i}-1}$, such that

$\displaystyle \mathcal{A}_2^{\vee} \simeq \mathbb{Z}/2[\xi_1, \xi_2, \dots ].$

The ${\xi_i}$ are defined as the dual elements to ${\mathrm{Sq}^{2^{i-1}} \mathrm{Sq}^{2^{i-2}} \dots \mathrm{Sq}^1}$ (with respect to the Serre-Cartan basis of ${\mathcal{A}_2}$).

We have a comodule structure map

$\displaystyle \phi: H_*(\mathbb{RP}^\infty; \mathbb{Z}/2) \rightarrow H_*(\mathbb{RP}^\infty; \mathbb{Z}/2) \otimes \mathcal{A}_2^{\vee}.$

Another way of phrasing the previous result is that the coefficient of ${b_1}$ in ${\phi(b_i)}$ is ${b_1 \otimes \zeta_k}$ if ${i = 2^k}$, and zero otherwise. That is,

$\displaystyle \phi(b_i) =\begin{cases} b_i \otimes 1 + \dots & \text{ if } i \neq 2^k, \forall k \\ b_i \otimes 1 + b_1 \otimes \zeta_k + \dots & \text{ if } i = 2^k. \end{cases}$

Here the dots signify terms that do not include ${b_1}$ or ${b_i}$.

2. The homology of ${MO}$, II

Let’s now finish working out the homology of ${MO}$ as a comodule over ${\mathcal{A}_2^{\vee}}$. We know that if ${T(\xi_1)}$ is the Thom space of the universal bundle on ${BO(1)}$, then ${T(\xi_1) \simeq BO(1) = \mathbb{RP}^\infty}$ (because the ${S^0}$-bundle over ${BO(1)}$ is contractible), and that the Thom class in ${T(\xi_1)}$ corresponds to the generator ${x \in H^1(\mathbb{RP}^\infty; \mathbb{Z}/2)}$.

In other words, we have that

$\displaystyle \widetilde{H}_*(T(\xi_1); \mathbb{Z}/2) \simeq \widetilde{H}_{*+1}(\mathbb{RP}^\infty) = \mathbb{Z}/2\left\{\beta_0, \beta_1, \dots\right\} ,$

where each ${\beta_i}$ is in degree ${i+1}$ and corresponds under the Thom isomorphism to ${b_i \in H_i(\mathbb{RP}^\infty)}$: equivalently, under the equivalence ${T(\zeta_1) \simeq \mathbb{RP}^\infty}$, ${\beta_i}$ corresponds to ${b_{i+1}}$. In particular,

$\displaystyle \widetilde{H}_{*}( MO(1)) \simeq \mathbb{Z}/2\left\{c_0, c_1, \dots\right\}$

where each ${c_i}$ is in degree ${i}$; this is because of the desuspension in ${MO(1)}$. Since ${H_*(BO(n))}$ is free on the appropriately ordered products of the ${b_i}$, and we have a commutative diagram (where all homology is with ${\mathbb{Z}/2}$-coefficients)

we find inductively that the vector space ${\widetilde{H}_*(MO(n))}$ is free on the products ${c_{i_1} \dots c_{i_n}}$ with ${i_0 \leq \dots \leq i_n}$. Observe that the desuspensions made in ${MO(n)}$ mean that we don’t have to change degrees anywhere. Moreover, ${c_0}$ becomes the unit becomes ${b_0}$ acts as the unit in ${H_*(BO; \mathbb{Z}/2)}$.

So, putting everything together, we find:

Proposition 5 We have an isomorphism of rings

$\displaystyle H_*(MO; \mathbb{Z}/2) \simeq \mathbb{Z}/2[c_1, c_2, \dots ].$

I’ve just been writing reduced homology for emphasis: that’s the natural thing for spectra, anyway.

The next goal is to figure out the action of ${\mathcal{A}_2^{\vee}}$ on this homology. The map ${\widetilde{H}_*(MO) \rightarrow \widetilde{H}_*(MO) \otimes \mathcal{A}_2^{\vee}}$ is just the map

$\displaystyle \pi_* MO \wedge H \mathbb{Z}/2 \rightarrow \pi_* MO \wedge H\mathbb{Z}/2 \wedge H \mathbb{Z}/2$

and it is in particular a morphism of rings. So we just need to figure out where each ${c_i}$ goes.

In fact, ${c_i}$ comes from ${H_*(MO(1)) \simeq \mathbb{Z}/2\left\{c_0, c_1, \dots\right\}}$ where each ${c_i}$ is in degree ${i}$: the element ${c_0}$ corresponds to ${1 \in H_*(MO)}$. By naturality, we might as well work out the coaction of ${\mathcal{A}_2^{\vee}}$ on ${H_*(MO(1))}$. Let’s work it out mod decomposables.

Proposition 6 The coaction ${\phi: H_*(MO(1)) \rightarrow H_*(MO(1)) \otimes \mathcal{A}_2^{\vee}}$ sends

$\displaystyle c_i \mapsto \begin{cases} c_i \otimes 1 + \dots & \text{if } i + 1 \neq 2^k, \forall k \\ c_i \otimes 1 + c_0 \otimes \zeta_k + \dots& \text{if } i + 1 = 2^k \end{cases} .$

The dots signify terms not involving ${c_i}$ or ${c_0}$.

Proof: To see this, we just have to remember that ${MO(1)}$ is a desuspension of ${\mathbb{RP}^\infty}$, and consequently ${H_*(MO(1))}$ is just a shift of ${\widetilde{H}_*(\mathbb{RP}^\infty)}$. Consequently, the ${c_i}$‘s are corresponding to the ${b_{i+1}}$‘s under the identification ${MO(1) \simeq \Sigma^{-1} \mathbb{RP}^\infty}$. If we keep this in mind, then the formula is a consequence of the previous analysis for ${\mathbb{RP}^\infty}$. $\Box$

Now, if we recall that ${c_0}$ is just the unit element of ${H_*(MO)}$ under the map ${H_*(MO(1)) \rightarrow H_*(MO)}$, then we find that the coaction ${\phi: H_*(MO) \rightarrow H_*(MO) \otimes \mathcal{A}_2^{\vee}}$ sends

$\displaystyle c_i \mapsto \begin{cases} c_i \otimes 1 + \dots & \text{if } i + 1 \neq 2^k, \forall k \\ c_i \otimes 1 + 1 \otimes \zeta_k + \dots& \text{if } i + 1 = 2^k \end{cases},$

where the ${\dots}$ signify decomposable elements in ${H_*(MO) \otimes \mathcal{A}_2^{\vee}}$.

The main result is:

Theorem 7 As an ${\mathcal{A}_2^{\vee}}$-comodule and an algebra, there is an isomorphism

$\displaystyle H_*(MO) \simeq \mathbb{Z}/2[d_i]_{i + 1 \neq 2^k} \otimes \mathcal{A}_2^{\vee}.$

When we analyze ${H_*(MU)}$ as a comodule, we will get something similar, but it will not be quite so simple.

Proof: The strategy is to produce a map of comodule-algebras ${H_*(MO) \rightarrow \mathbb{Z}/2[d_i]_{i + 1 \neq 2^k} \otimes \mathcal{A}_2^{\vee}}$. We can do this very easily: it’s the same thing as a map of plain algebras

$\displaystyle \psi: H_*(MO) \rightarrow \mathbb{Z}/2[d_i]_{i + 1 \neq 2^k}$

because the tensor product ${\mathbb{Z}/2[d_i]_{i + 1 \neq 2^k} \otimes \mathcal{A}_2^{\vee} }$ 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 ${\psi(c_i) = d_i}$ if ${i + 1 \neq 2^k}$ and zero otherwise.

The map of comodule-algebras is then obtained via the composition

$\displaystyle \Psi: H_*(MO) \rightarrow H_*(MO) \otimes \mathcal{A}_2^{\vee} \stackrel{\psi \otimes 1}{\rightarrow} \mathbb{Z}/2[d_i]_{i+1 \neq 2^k} \otimes \mathcal{A}_2^{\vee},$

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 ${\mathcal{A}_2^{\vee} = \mathbb{Z}/2[\zeta_1, \zeta_2, \dots ]}$.

To show that the composite map ${\Psi}$ is an isomorphism, we’re going to have to use a commonly used technique: we’ll look at indecomposable elements. In dimension ${i}$, the indecomposable elements in ${H_*(MO)}$ is spanned the image of ${c_i}$. We know, moreover, from our earlier computations that modulo decomposables,

$\displaystyle \Psi(c_i) = \begin{cases} c_i \otimes 1 & \text{if } i+1 \neq 2^k \forall k \\ 1 \otimes \zeta_k & \text{if } i + 1 = 2^k \end{cases}.$

This is, in either case, indecomposable in degree ${i}$ in ${\mathbb{Z}/2[d_i]_{i + 1 \neq 2^k} \otimes \mathcal{A}_2^{\vee}}$. In particular, the map ${\Psi}$ sends indecomposables to indecomposables, so by induction on the degree we find that ${\Psi}$ is a surjection. Since the domain and the codomain have the same dimension in each degree, ${\Psi}$ must be an isomorphism. $\Box$

3. The homotopy of ${MO}$

In the previous sections, we worked out completely the homology of ${MO}$ as a comodule over the coalgebra ${\mathcal{A}_2^{\vee}}$, as well as an algebra in its own right. We know want to work out the spectrum ${MO}$, and prove the theorems stated at the start of the blog post.

So, we want to show that ${MO}$ is a wedge of Eilenberg-MacLane spectra.

Proposition 8 Suppose ${X}$ is a connective spectrum, whose homotopy groups ${\pi_* X}$ are all 2-torsion. Suppose ${H_*(X; \mathbb{Z}/2)}$ is a co-induced module over ${\mathcal{A}_2^{\vee}}$, i.e. ${H_*(X; \mathbb{Z}/2) = M \otimes \mathcal{A}_2^{\vee}}$ for a graded ${\mathbb{Z}/2}$-vector space ${M}$ whose graded pieces are finite-dimensional.Choose a basis ${b_{i,k}}$ of the graded piece ${M_k}$. Then we have a decomposition of spectra

$\displaystyle X \simeq \bigvee_{i, k} H\mathbb{Z}/2[k].$

Proof: The strategy will be to produce a map ${ X \rightarrow \bigvee_{i, k} H\mathbb{Z}/2[k] \simeq \prod_{i,k} H \mathbb{Z}/2[k]}$ which is an isomorphism on ${\mathbb{Z}/2}$-homology, and which will then be forced to be an equivalence.

In fact, we find that ${H^*(X; \mathbb{Z}/2)}$ is free as a module over the Steenrod algebra, ${H^*(X; \mathbb{Z}/2) \simeq M^{\vee} \otimes \mathcal{A}_2}$. The dual basis elements ${b^{\vee}_{i, k}}$ are cohomology classes in dimension ${i}$ which define a map

$\displaystyle X \rightarrow \bigvee_{i,k} H\mathbb{Z}/2[k].$

This map is an isomorphism on cohomology by construction: the cohomology of an Eilenberg-MacLane spectrum is free on ${\mathcal{A}_2}$.

Dualizing, we find that it is an isomorphism on ${\mathbb{Z}/2}$-homology. It follows that the cofiber, which is a connective spectrum with 2-torsion homotopy groups, has trivial ${\mathbb{Z}/2}$-homology. This means that the cofiber is contractible: for, if its first nonzero homotopy group were in degree ${i}$, then we would have ${H_i(\mathrm{cof}; \mathbb{Z}/2) = \pi_i(\mathrm{cof}) \otimes \mathbb{Z}/2 \neq 0}$. $\Box$

So, in particular, applying this to ${MO}$, we find that ${MO}$ is a wedge of shifts of ${H \mathbb{Z}/2}$. This means in particular that the Hurewicz map

$\displaystyle \pi_* MO \rightarrow H_*(MO) \simeq \mathbb{Z}/2[d_i]_{i + 1 \neq 2^k} \otimes \mathcal{A}_2^{\vee},$

is an injection, and that its image must consist of the primitive elements: these are given by

$\displaystyle \pi_* MO \simeq \mathbb{Z}/2[d_i]_{i + 1 \neq 2^k} .$

This completes the proof of Thom’s theorem.