Our goal is now to return to topology, and in particular to study the formal group law of the universal complex-oriented theory ${MU}$ (complex cobordism). As we computed using the Adams spectral sequence,

$\displaystyle \pi_* MU \simeq \mathbb{Z}[x_1, x_2, \dots ] , \quad \deg x_i = 2i.$

This is the Lazard ring, by the computations of the previous couple of posts. On the other hand, it is not at all clear that the map

$\displaystyle L \rightarrow \pi_* MU$

classifying the formal group law over ${\pi_* MU}$ (arising from the complex orientation) is actually an isomorphism: in other words, that the formal group law of ${MU}$ is the universal one. The fact that it is in fact an isomorphism is the content of Quillen’s theorem, which will be proved in this post.

1. The Hurewicz map

There is a Hurewicz map

$\displaystyle \pi_* MU \rightarrow H_* (MU) = \mathbb{Z}[b_1, b_2, \dots ].$

In particular, the map ${L \rightarrow \pi_* MU}$ leads to a map ${L \rightarrow \mathbb{Z}[b_1, \dots ]}$ classifying a formal group law over ${\mathbb{Z}[b_1, \dots ]}$. This looks suspiciously like the map ${L \rightarrow \mathbb{Z}[b_1, \dots]}$ earlier, and in fact it is the same thing. Once we know this, we’ll be able to get a good handle on the formal group law for ${\pi_* MU}$, because ${\pi_* MU}$ and ${H_*MU}$ are close.

So we have:

Claim: The formal group law on ${H_*(MU; \mathbb{Z})}$ classified by the map ${L \rightarrow \pi_* MU \rightarrow H_*(MU)}$ is the formal group law ${\exp(\exp^{-1}(x) + \exp^{-1}(y))}$ where ${\exp(x) = \sum b_i x^{i+1}}$.

We will in fact prove something more general. Let ${E}$ be a complex-oriented ring spectrum. Then the smash product ${E \wedge MU}$ is a complex-oriented ring spectrum in two different ways:

1. There is a complex orientation coming from the ring-spectra morphism ${E \rightarrow E \wedge MU}$ and the complex orientation of ${E}$.
2. There is a complex orientation coming from the ring-spectra morphism ${MU \rightarrow E \wedge MU}$ and the complex orientation of ${MU}$.

Each of these classify two different formal group laws ${f_1, f_2}$ over the ring ${\pi_* E \wedge MU = \pi_* E [b_1, b_2, \dots ]}$. The formal group law ${f_1}$ is just the formal group law of ${E}$. The formal group law ${f_2}$ is a little different.

Claim: the formal group law ${f_2}$ over ${\pi_* E [b_1, b_2, \dots ]}$ is obtained from ${f_1}$ by the “change of coordinates” ${\exp(x) = \sum b_i x^{i+1}}$.

In particular, smashing a complex-oriented spectrum with ${MU}$ can be thought of as a universal change of coordinates of the associated formal group law.

This is, of course, a generalization of the previous claim: in that case, ${E = H \mathbb{Z}}$ and the formal group law of ${H \mathbb{Z}}$ is the additive one.

2. The proof of the claim

Let’s now try to prove this claim. Let ${E}$ be a complex-oriented ring spectrum. We know then that

$\displaystyle (E \wedge MU)^*(\mathbb{CP}^{\infty}) = \pi_* E \wedge MU [[t]] = \pi_* E [b_1, b_2, \dots] [[t]]$

for any choice ${t}$ of complex orientation of ${E \wedge MU}$ (that is, an element of ${(E \wedge MU)^2(\mathbb{CP}^\infty)}$). Of course, we have two such choices of ${t}$:

1. We have ${t_E}$, which comes from the orientation of ${E}$.
2. We have ${t_{MU}}$, which comes from the orientation of ${MU}$.

The two formal group laws on ${\pi_* E[b_1, b_2, \dots ]}$ are both manifestations of the map

$\displaystyle (E \wedge MU)^*(\mathbb{CP}^{\infty}) \rightarrow (E \wedge MU)^*(\mathbb{CP}^{\infty} \times \mathbb{CP}^\infty),$

and just depend on the choice of complex orientation ${t}$ to be expressed in terms of a formal power series.

So, in other words, our goal is to express ${t_{MU}}$ in terms of ${t_E}$. Our goal is to prove that

$\displaystyle t_{MU} = \sum b_i t_E^{i+1} \quad (b_0 = 1).$

If we prove this, it will follow that the two formal group laws over ${\pi_* E \wedge MU}$ are connected by the “change of coordinates” ${\sum b_i t^{i+1}}$, which will prove the claim.

How might we evaluate ${t_{MU}}$ in terms of ${t_E}$? The strategy is to pair against classes in ${(E \wedge MU)_* (\mathbb{CP}^\infty}$. This is spanned by classes ${\beta_i, 0 \leq i < \infty}$. We choose the ${\beta_i}$ as coming the complex orientation of ${E}$: that is, they are the image of the basis of ${E_*(\mathbb{CP}^\infty)}$ under the map

$\displaystyle E_*(\mathbb{CP}^\infty) \rightarrow (E \wedge MU)_* (\mathbb{CP}^\infty).$

In other words, ${\left\{\beta_i\right\}}$ is the dual basis to ${\left\{t_E^i\right\}}$.

So, let’s pair ${t_{MU}}$ against the classes ${\beta_i}$. First, we should recall what “pairing” means. The class ${t_{MU}}$ is represented by the map

$\displaystyle t_{MU}: \mathbb{CP}^\infty \rightarrow MU \rightarrow E \wedge MU$

of degree two. Each ${\beta_i}$ is represented by a map

$\displaystyle \beta_i: S \rightarrow \mathbb{CP}^\infty \wedge E \rightarrow \mathbb{CP}^\infty \wedge E \wedge MU.$

In order to do the pairing, we have to form

$\displaystyle S \stackrel{\beta_i}{\rightarrow} \mathbb{CP}^\infty \wedge E \wedge MU \stackrel{t_{MU}}{\rightarrow} \mathbb{CP}^\infty \wedge E \wedge E \wedge MU \wedge MU \rightarrow \mathbb{CP}^\infty \wedge E \wedge MU$

where the last map is multiplication. Because both ${\beta_i}$ and ${t_{MU}}$ came from maps ${S \rightarrow \mathbb{CP}^\infty \wedge E}$ and ${\mathbb{CP}^\infty \rightarrow MU}$, we find that the big composite is equal to

$\displaystyle S \stackrel{\beta_i}{\rightarrow} \mathbb{CP}^\infty \wedge E \stackrel{t_{MU}}{\rightarrow} MU \wedge E,$

where ${\beta_i}$ is identified with the analogous element in ${E_*(\mathbb{CP}^\infty)}$ and ${t_{MU}}$ with the analogous element in ${MU^*(\mathbb{CP}^\infty)}$.

But this is just saying that $(t_{MU}, \beta_i)$ is the image of ${\beta_i}$ under the map (of degree two) ${\mathbb{CP}^\infty \rightarrow MU}$ in ${E}$-homology. In other words,

$\displaystyle (t_{MU}, \beta_i) = b_{i-1},$

because the ${\left\{b_i\right\}}$ are precisely the image of the basis of ${(E)_* (MU(1))}$ in ${(E )_* (MU)}$. Since ${\left\{\beta_i\right\}}$ is the dual basis in ${(MU \wedge E)_*(\mathbb{CP}^\infty)}$ to ${\left\{t_E^{i}\right\}}$, we get

$\displaystyle t_{MU} = \sum_{i \geq 1} b_{i-1}t_E^{i} .$

This is precisely the claim about the generator ${t_{MU}}$.

As we said before, this means that the formal group law of ${E \wedge MU}$ in ${MU}$-coordinates is just a twist of that of the FGL of ${E}$, under the power series ${\sum_{i \geq 0} b_i x^{i+1}}$.

3. Quillen’s theorem

Now we are almost at the end of the proof of Quillen’s theorem. What do we know?

1. ${\pi_* MU}$ is a polynomial ring ${\mathbb{Z}[x_1, x_2, \dots]}$. There is a map ${L \rightarrow \pi_* MU}$ classifying the formal group law of ${MU}$.
2. There is a Hurewicz map ${\pi_* MU \rightarrow H_*(MU; \mathbb{Z})}$. The map ${L \rightarrow \pi_* MU \rightarrow H_*(MU)}$ classifies the formal group law ${\exp( \exp^{-1}(x) + \exp^{-1}(y))}$ where ${\exp (x) = \sum b_i x^{i+1}}$.
3. We have studied the map ${L \rightarrow H_*(MU)}$ before: in particular, we have seen that it is an injection, and determined that, on indecomposables, the map

$\displaystyle Q_m(L) \rightarrow Q_m(H_*(MU))$

(where ${Q}$ denotes indecomposables) is an isomorphism if ${m +1}$ is not a power of a prime, and is given by multiplication by ${p: \mathbb{Z} \rightarrow \mathbb{Z}}$ if ${m+1 = p^f}$.

Observe now that the map ${\pi_* MU \rightarrow H_*(MU)}$ is injective, and its image contains the image of the Lazard ring ${L}$. If we show, conversely, that the image of ${\pi_* MU \rightarrow H_*(MU)}$ is contained in the image of the Lazard ring, then we will be done. We can check this on indecomposables.

In particular, we are reduced to showing that if ${n + 1 = p^f}$ for a prime ${p}$, then the map ${\pi_* MU \rightarrow H_*(MU)}$ induces

$\displaystyle \mathbb{Z} \simeq Q_n(\pi_*(MU)) \rightarrow Q_n( H_*(MU)) \simeq \mathbb{Z},$

whose image is contained in ${p \mathbb{Z}}$.

Proof: It is equivalent to say that the mod ${p}$ Hurewicz homomorphism

$\displaystyle Q_n(\pi_*(MU)) \rightarrow Q_n( H_*(MU; \mathbb{Z}/p))$

is zero. But this actually follows from the Adams spectral sequence for ${MU}$ done a few blog posts back. In fact, we saw that in the ASS converging to ${\pi_* MU}$, the indecomposables in degree ${p^f - 1}$ had Adams filtration (i.e., ${s}$ in the spectral sequence) at least one. For a class to have Adams filtration ${\geq 1}$ is precisely to say that it is annihilated by the Hurewicz map—this is a consequence of how one sets up the spectral sequence with Adams resolutions. $\Box$

So we can finally state:

Theorem 1 (Quillen) The formal group law of ${MU}$ is the universal one.

It’s sort of striking, actually, how computational the proof of Quillen’s theorem was, when the result itself is very conceptual: that the universal complex-oriented theory should correspond to the universal formal group law. However, there appears to be no non-computational proof, even though, for instance, a construction of ${MU}$ itself from formal group laws would be extremely awesome.