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:

  1. 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.

  2. 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.
  3. 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:

  1. 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.
  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

  1. {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.
  2. 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.

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