The formal group law of unoriented cobordism

Before moving on, I’d like to work out the analog for real-oriented cohomology theories (which is tangential to the rest of the story, though). This is considerably less interesting, but perhaps it’s a toy example of the ideas explained in the last few posts without the full-blown machinery of the Adams spectral sequence and so forth.

So, let’s state what the analogous ideas are in the real-oriented context:

A real-oriented ring spectrum ${E}$ is a ring spectrum together with a functorial, multiplicative choice of Thom classes for real vector bundles; equivalently, there is a morphism of ring spectra
$\displaystyle MO \rightarrow E.$

That is, the universal real-oriented ring spectrum is unoriented cobordism, whose homotopy groups can be completely computed.
If ${E}$ is real-oriented, then ${\pi_* E}$ is a ${\mathbb{Z}/2}$ -vector space, and all the usual computations of ${H_*( BO; \mathbb{Z}/2)}$ and so forth work just fine for ${E}$ , and there is a theory of Stiefel-Whitney classes in ${E}$ -cohomology.
Given a real-oriented spectrum ${E}$ , we thus have ${E^*(\mathbb{RP}^\infty) = \pi_* E [[t]]}$ where ${t}$ is the Stiefel-Whitney class of the tautological bundle. Similiarly, ${E^*(\mathbb{RP}^\infty \times \mathbb{RP}^\infty) = \pi_* E [[t_1, t_2]]}$ .

Since ${\mathbb{RP}^\infty}$ is the classifying space for real line bundles, there is a monoidal product

$\displaystyle \mathbb{RP}^\infty \times \mathbb{RP}^\infty \rightarrow \mathbb{RP}^\infty,$

classifying the tensor product of line bundles. As before, this means that we can extract a formal group law over ${\pi_* E}$ . This formal group law ${f(\cdot, \cdot) \in \pi_* E [[x, y]]}$ has the property that if ${\mathcal{L}_1, \mathcal{L}_2}$ are two line bundles over a finite-dimensional space ${X}$ , then in ${E^*(X)}$ ,

$\displaystyle w_1( \mathcal{L}_1 \otimes \mathcal{L}_2) = f( w_1(\mathcal{L}_1), w_1(\mathcal{L}_2)).$

So far everything has been analogous to the complex-oriented case, but there is an extra feature here which changes the picture drastically. Namely, when one works with realline bundles ${\mathcal{L}}$ , we have that ${\mathcal{L}^{\otimes 2}}$ is always trivial. (The analog is very false for complex line bundles.) This means that the formal group law ${f}$ must satisfy

$\displaystyle f(x, x) = 0.$

This is not satisfied by, say, the multiplicative formal group law. (And in fact, ${KO}$ -theory—the natural candidate for this—is not real-oriented, only spin-oriented.)

With this in mind, the analog of Quillen’s theorem for ${MU}$ becomes:

Theorem 1 (Quillen) The formal group law for ${MO}$ is the universal formal group law over a satisfying ${f(x, x) = 0}$ .

The condition that ${f(x, x) = 0}$ implies, for instance, that the base ring of such a formal group law is a ${\mathbb{Z}/2}$ -algebra. It is equivalent to the condition that ${f}$ have infinite height, and implies in particular that ${f}$ is isomorphic to the additive formal group law. This is why the theory of ${MU}$ is so much richer.

How can we prove such a result? Let ${R}$ be the ring classifying the universal formal group law ${f}$ satisfying ${f(x, x) =0}$ ; then there is a map

$\displaystyle R \rightarrow \pi_* E$

for every real-oriented spectrum ${E}$ . We would like to prove that it is an isomorphism when ${E = MO}$ . Fortunately, we already know that

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

and so we will determine what ${R}$ looks like, in analogy with the analysis of the Lazard ring.

1. The universal ring

There is a map

$\displaystyle R \rightarrow \mathbb{Z}/2[b_1, b_2, \dots ]$

classifying the formal group law ${\exp( \exp^{-1}(x) + \exp^{-1}(y))}$ over ${\mathbb{Z}/2[b_1, b_2, \dots ]}$ ; since we have introduced ${\mathbb{Z}/2}$ -coefficients this formal group law satisfies ${f(x, x) =0 }$ . In order to understand ${R}$ , we will do something familiar: we will study the map on indecomposables induced.

As before, this is equivalent to studying appropriately graded formal group laws over rings of the form ${\mathbb{Z}/2 \oplus A}$ , for ${A}$ a ${\mathbb{Z}/2}$ -module. In other words, we are interested in, for each ${n}$ , polynomials of the form

$\displaystyle x + y + \sum_{i + j = n+1} a_{i,j}x^i y^j, \quad a_{i,j} \in A$

which define formal group laws over ${\mathbb{Z}/2 \oplus A}$ of infinite height.

Anyway, the condition that this be a formal group law implies that the polynomial ${\Gamma(x,y) = \sum_{i+j = n+1} a_{i,j} x^i y^j}$ is a symmetric 2-cocycle as in the sense of the previous posts. In particular, we find:

If ${n + 1}$ is a power of ${2}$ , then ${\Gamma(x,y) = a \frac{1}{2} ( (x+y)^{n+1} - x^{n+1} - y^{n+1} )}$ for some ${a \in A}$ . Here the polynomial ${\frac{1}{2}( (x+y)^{n+1} - x^{n+1} - y^{n+1} ) }$ has ${\mathbb{Z}}$ -coefficients and is regarded as reduced mod 2.
If ${n+1}$ is not a power of ${2}$ , then ${\Gamma(x,y) = a ( (x+y)^{n+1} - x^{n+1} - y^{n+1} )}$ for some ${a \in A}$ .

This is a consequence of the “symmetric 2-cocycle lemma” and the fact that all primes other than ${2}$ are invertible.

We have also seen that the FGL which arise from a change-of-coordinates in this way are those where ${\Gamma(x,y)}$ is a multiple of ${(x +y)^{n+1} - x^{n+1} - y^{n+1}}$ (this follows from a direct computation).

But we have one more condition: we want that ${\Gamma(x, x) = 0}$ if our formal group law is of infinite height. In case 2 above, this is automatically satisfied. In case 1, it is not: we get ${\Gamma(x, x) = (2^n - 1)ax ^{n+1}}$ . It follows that in case 1, ${a}$ must be zero, and there are nonontrivial formal group laws of that form.

Putting it together, we find:

Proposition 2

${Q_n(R) = \mathbb{Z}/2}$ if ${n+1}$ is not a power of ${2}$ and the map ${Q_n(R) \rightarrow Q_n( \mathbb{Z}/2[b_1, b_2, \dots ])}$ is an isomorphism.

If ${n+1}$ is a power of ${2}$ , then ${Q_n(R) =0 }$ .

Consequently, we find that

$\displaystyle R = \mathbb{Z}/2[x_i]_{i + 1 \neq 2^k},$

and the “algebraic Hurewicz map” ${R \rightarrow \mathbb{Z}/2[b_1, b_2, \dots ]}$ is just given by adding new polynomial generators.

2. Putting it together

Now there’s not much left to do. We’ve given a complete description of the ring ${R}$ classifying formal group laws of infinite height at ${2}$ , and we know how ${\pi_* MO}$ looks—exactly the same. The thing to check is that the map

$\displaystyle R \rightarrow \pi_* MO$

is actually an isomorphism.

As before, we can study this map by studying the composite

$\displaystyle R \rightarrow \pi_*MO \rightarrow H_*(MO ; \mathbb{Z}/2),$

and using the same arguments as the previous time, we can argue that the FGL on ${H_*(MO; \mathbb{Z}/2)}$ classified by it is in fact ${\exp( \exp^{-1}(x) + \exp^{-1}(y))}$ . But since the image of ${R}$ in ${H_*(MO; \mathbb{Z}/2)}$ is the same as that of ${\pi_* MO}$ , and since ${\pi_* MO }$ is isomorphic as a graded vector space to ${R}$ , we find that the map ${R \rightarrow \pi_* MO}$ is an isomorphism.

Climbing Mount Bourbaki