Complex oriented cohomology theories$MU(n)$ and $MU$

# 2 $MU(n)$ and $MU$

Definition 2.1

We define $MU(n)$ to be the Thom space of the tautological vector bundle of $BU(n)$.

We define $MU$ to be the Thom spectrum of the tautological virtual vector bundle of $BU$. Equivalently,

$MU = \operatorname*{colim}_n \Sigma ^{-2n} \Sigma ^\infty MU(n).$

The direct sum map $BU \times BU \to BU$ induces a map $MU \wedge MU \to MU$, which turns $MU$ into a ring spectrum (and in fact an $\mathbb {E}_\infty$-ring spectrum).

The crucial insight of Thom was that this spectrum is closely related to cobordism theory.
Theorem 2.2 (Pontryagin–Thom)

Let $X$ be a space. Then $MU_*(X_+)$ admits the following description:

1. A class in $MU_d(X_+)$ is represented by a $d$-dimensional stably almost complex manifold $M$ and a map $f: M \to X$. Addition is given by disjoint union.

2. Two classes $f_1, f_2$ are equivalent if there is a cobordism $g: N \to X$ between them.

Proof
[Proof idea] Given a class in $MU_d(X_+)$, I explain how to get such a map $f: M \to X$. By the Whitney embedding theorem.

A map $S^d \to MU \wedge X_+$ factors through a map $S^d \to \Sigma ^{-2n} MU(n) \wedge X_+$ for some $n$, and is thus given a map of spaces $\phi : S^{d + 2n} \to MU(n) \wedge X_+$, increasing $n$ if necessary (since $\Sigma ^{2k} MU(n) \hookrightarrow MU(n + k)$). Recall that

$MU(n) \wedge X_+ = MU(n) \times X / \{ *\} \times X,$

where the point on the right is point at infinity. $MU(n)$ also contains the “zero section” isomorphic to $BU(n)$, and the normal bundle of the zero section is the tautological bundle of $BU(n)$. Generically, we can choose $\phi$ to intersect transversely in a way that $\phi ^{-1}(BU(n) \times X)$ is a codimension $2n$ submanifold of $S^{d + 2n}$, and the normal bundle, being pulled back from $BU(n)$, has an almost complex structure. This gives a stably almost complex manifold of dimension $d$ with a map to $X$. A homotopy of $\phi$ gives a cobordism between such maps.

Proof

So in particular, $MU_* = \pi _* MU$ is the complex cobordism group. Remarkably, these groups are entirely computable.

Proposition 2.3
\begin{aligned} H_*(MU) & = \mathbb {Z}[b_1, b_2, b_3, \ldots ]\\ \pi _*(MU) & = \mathbb {Z}[m_1, m_2, m_3, \ldots ] \end{aligned}

where $|m_i| = |b_i| = 2i$. The Hurewicz map $\pi _*(MU) \to H_*(MU)$ is injective but not surjective.

Proof
[Proof sketch] Since the tautological vector bundle of $BU$ is oriented, the first part follows from the Thom isomorphism theorem and the standard calculation of the homology of $BU$. The homotopy groups follow from a more involved Adams spectral sequence calculation.
Proof