# 4 Formal group laws

For the remainder of the talk, I wish to use a bit more algebraic geometry language. We assume $E$ is a commutative complex oriented ring spectrum. If $X$ is a CW complex, then $E^{2*}X_+$ is a ring. We restrict to the evenly graded parts so that it is actually commutative. If one is concerned, one can take into account all degrees and forgo the algebro-geometric language, but doing so gains us nothing (this works out because the spaces we care about only have even cells).

To be honest mathematicians, we should consider $E^{2*} X_+$ as a topological ring (or a pro-ring), with the topology given by

$E^{2*} X_+ = \lim _{A \subseteq X \text{ finite}} E^{2*} A_+$For example,

$E^{2*} \mathbb {CP}^\infty _+ = \lim _n E^{2*} \mathbb {CP}^n = E^{2*}[\! [u]\! ],$where the power series ring has the usual topology.

Since we are doing algebraic geometry, we are supposed to apply the functor $\operatorname{Spec}$ to rings. Actually, since we have topological rings, we should apply $\operatorname{Spf}$ instead, which remembers the topology, and end up with a *formal scheme*. If you don't know about $\operatorname{Spf}$, you can pretend it is $\operatorname{Spec}$ instead. The upshot is that the functor

is a *covariant* functor in $X$. Moreover, if we restrict to spaces for which $E^*X_+$ is a free $E^*$-module, such as $\mathbb {CP}^n$ and complex oriented cohomology theories, this functor is symmetric monoidal by Künneth's theorem.

For $\mathbb {CP}^\infty$, we have

$\mathbb {CP}^\infty _E = \operatorname{Spf}E^{2*}[\! [u]\! ].$This as an infinitesimal neighbourhood of $0 \in \mathbb {A}^1$ over $\operatorname{Spec}E^{2*}$, which we denote $\hat{\mathbb {A}}^1$. Our first conclusion is thus

A complex orientation of $E$ gives an isomorphism

$\mathbb {CP}^\infty _E = \hat{\mathbb {A}}^1.$ The fact that $E$ is complex orienta*ble* tells us $\mathbb {CP}^\infty _E$ is abstractly isomorphic to $\hat{\mathbb {A}}^1$, and a complex orientation is a choice of isomorphism.

Now recall that $X \mapsto X_E$ is symmetric monoidal. Moreover, $\mathbb {CP}^\infty$ has the structure of an abelian group (in the homotopy category), with the map $\otimes : \mathbb {CP}^\infty \times \mathbb {CP}^\infty \to \mathbb {CP}^\infty$ classifying the tensor product of line bundles. This turns $\mathbb {CP}^\infty _E$ into a (formal) group scheme.

A formal group is a commutative formal group scheme whose underlying scheme is (locally) isomorphic $\hat{\mathbb {A}}^1$.

A formal group law is a commutative group scheme where the underlying scheme is *equipped with* an isomorphism with $\hat{\mathbb {A}}^1$.

By convention, an isomorphism of formal group laws is an isomorphism of the underlying formal groups (that is, it is not required to act as the identity on $\hat{\mathbb {A}}^1$, or else they are extremely boring).

*ble*, then $\mathbb {CP}^\infty _E$ is a formal group. If it is complex orient

*ed*, then $\mathbb {CP}^\infty _E$ is given the structure of a formal group

*law*. Different choices of complex orientations give different but isomorphic formal group laws.

Let us unwrap what it means to be a formal group law. Let $R$ be a ring. A formal group law is a map

$\operatorname{Spf}R[\! [x]\! ] \times \operatorname{Spf}R[\! [x]\! ] \to \operatorname{Spf}R[\! [x]\! ]$satisfying certain properties. Undoing the $\operatorname{Spf}$ gives us a continuous map of $R$-algebras

$R[\! [x]\! ] \to R[\! [x, y]\! ].$This is uniquely determined by the value of $x$. Call this $x +_F y$, which is a power series in $x$ and $y$. The property of being a commutative group is equivalent to the conditions

$\begin{aligned} x +_F y & = y +_F x\\ x +_F 0 & = 0\\ (x +_F y) +_F z & = x +_F (y +_F z). \end{aligned}$An isomorphism of formal group laws is given by an automorphism of $R[\! [x]\! ]$ that sends one formal group law to the other. Again an automorphism $R[\! [x]\! ] \to R[\! [x]\! ]$ is uniquely specified by the image of $x$, say $f(x)$, and an isomorphism between $+_F$ and $+_G$ is an invertible $f$ such that

$f(x +_F y) = f(x) +_G f(y).$A formal group is then a formal group law up to isomorphism.

Note that if $f: R \to S$ is a map of rings, then a formal group law over $R$ induces a formal group law over $S$ by applying $f$ to the coefficients of the power series. Since we are thinking in terms of schemes, we call this “pulling back” the formal group law from $\operatorname{Spec}R$ to $\operatorname{Spec}S$.

There is a universal formal group law. That is, there is a ring $L$ with a formal group law $\hat{\mathbb {G}}_L$ on $L$ such that for any other formal group law $\hat{\mathbb {G}}$ on a ring $R$, there is a unique map $f: L \to R$ such that $\hat{\mathbb {G}} = f^* \hat{\mathbb {G}}_L$. Moreover, $L \cong \mathbb {Z}[\ell _1, \ell _2, \ell _3, \ldots ]$.

Recall that $MU$ is the universal complex oriented cohomology theory, and a complex oriented cohomology theory has a canonical formal group law.

$\mathbb {CP}^\infty _{MU}$ is the universal formal group law.

In general, the formal group of a complex orientable cohomology theory captures a lot of important information about the theory. It also allows us to perform some nice computations:

If $E$ is complex oriented and $F$ is any ring spectrum, then $E \wedge F$ is also complex oriented, and the formal group law on $E \wedge F$ is pulled back from that of $E$.

Thus, if both $E$ and $F$ are complex oriented, then we get two complex orientations of $E \wedge F$, hence two formal group laws. However, since the formal group of a complex orientable cohomology theory is well-defined, these two formal group laws must be isomorphic.

For example, take $E = H\mathbb {Z}$ and $F = KU$. The respective formal group laws are $\hat{\mathbb {G}}_a$ and $\hat{\mathbb {G}}_m$. We then know that $\pi _*(H\mathbb {Z}\wedge KU)$ is a ring on which $\hat{\mathbb {G}}_a$ and $\hat{\mathbb {G}}_m$ are isomorphic. Standard theory of formal group laws tells us this is possible only if the ring is rational (since one has height $\infty$ and the other has height $1$). So $H_*KU = \pi _*(H\mathbb {Z}\wedge KU)$ is rational.

Now consider the rationalization map $KU \to KU_\mathbb {Q}$. Since $H_* KU$ is already rational, this map is an isomorphism on integral homology. However, $\pi _* KU$ is definitely not rational, so the map is *not* an isomorphism on homotopy groups. Thus the cofiber of this map has trivial integral homology, but is non-contractible.