1 Thom spaces

Recall from your Algebraic Topology education that if $\xi : V \to X$ is an oriented vector bundle of rank $d$ , then we have isomorphisms

\begin{aligned} H_*(X_+) \cong H_{* + d}(V, V \setminus X),\\ H^*(X_+) \cong H^{* + d}(V, V \setminus X). \end{aligned}

called the Thom isomorphism. Later, as we grew up, we learnt that relative homology is just the homology of the cofiber in disguise. As such, we define

Definition 1.1

Let $\xi : V \to X$ be a vector bundle. The Thom space of $\xi$ , written $\mathrm{Th}'(\xi )$ , is the homotopy pushout ¹

$\begin{useimager} \[ \begin{tikzcd} V \setminus X \ar[r] \ar[d] & V \ar[d]\\ * \ar[r] & \Th'(\xi) \end{tikzcd}. \] \end{useimager}$

This makes $\mathrm{Th}'(\xi )$ a based space.

More concretely, pick a metric on $\xi$ arbitrarily. We can then define the sphere and disk bundles

\begin{aligned} S(V) & = \{ v \in V: \| v\| = 1\} ,\\ D(V) & = \{ v \in V: \| v\| \leq 1\} . \end{aligned}

Then

\mathrm{Th}'(\xi ) = D(V)/S(V).

Example 1.2

If $\xi$ is trivial of rank $d$ , then $\mathrm{Th}'(\xi ) = \Sigma ^d X_+$ .

Thus, we should think of the Thom space as a twisted suspension of

X

, where the twisting is specified by a vector bundle.

The construction of Thom spaces has two important properties, both of which are straightforward to verify:

It is functorial along pullbacks — if $f: X \to Y$ is a map and $\xi : V \to Y$ is a vector bundle, then there is a natural map
$\mathrm{Th}(f): \mathrm{Th}(f^* \xi ) \to \mathrm{Th}(\xi ).$
It is monoidal — if $\xi : V \to X$ and $\zeta : W \to Y$ are vector bundles and $\xi \boxplus \zeta : V \boxplus X \to X \times Y$ is the external direct sum. Then
$\mathrm{Th}'(\xi \boxplus \zeta ) \cong \mathrm{Th}'(\xi ) \wedge \mathrm{Th}'(\zeta ).$

Corollary 1.3

$\mathrm{Th}'(\xi \oplus \mathbb {R}) = \Sigma \mathrm{Th}'(\xi )$ .

Proof

□

\xi \oplus \mathbb {R}= \xi \boxplus (\mathrm{triv}: \mathbb {R}\to *)

and

\mathrm{Th}'(\mathrm{triv}) = S^1

Proof

□

This leads to the following natural definition:

Definition 1.4

If $\xi : V \to X$ is a rank $d$ vector bundle, then the Thom spectrum of $\xi$ is

\mathrm{Th}(\xi ) = \Sigma ^{-d} \Sigma ^\infty \mathrm{Th}'(\xi ).

This definition is designed so that

\mathrm{Th}(\xi \oplus \mathbb {R}) = \mathrm{Th}(\xi )

, and hence the Thom spectrum can be defined for virtual vector bundles as well. While we'll mostly state things for vector bundles, everything we say applies equally to virtual vector bundles. The monoidality property still holds for the Thom spectrum:

\mathrm{Th}(\xi \boxplus \zeta ) \cong \mathrm{Th}(\xi ) \wedge \mathrm{Th}(\zeta ).

With the Thom spectrum, we can rephrase the Thom isomorphism as saying

Theorem 1.5 (Thom)

If $\xi : V \to X$ is an oriented vector bundle, then we have a natural isomorphism of $H\mathbb {Z}$ -modules

H\mathbb {Z}\wedge \mathrm{Th}(\xi ) \simeq H\mathbb {Z}\wedge X_+.

If we think of Thom spectra as twisted suspensions, this says tensoring with

H\mathbb {Z}

untwists it.

The classical statement of the Thom isomorphism theorem is a bit more specific. It says the isomorphism is given by capping or cupping with a Thom class $u \in H^*(\mathrm{Th}(\xi ))$ . Usually, the cup product is induced by pulling back along the diagonal. However, the cup product we want here takes the form

\smile : H^*(\mathrm{Th}(\xi )) \otimes H^*(X) \to H^*(\mathrm{Th}(\xi )).

so that $u \smile \, \cdot$ gives an isomorphism. This cup product is given by the Thom diagonal.

Definition 1.6

Let $\xi : V \to X$ be a vector bundle. The Thom diagonal is a map

\Delta : \mathrm{Th}(\xi ) \to \mathrm{Th}(\xi ) \wedge X_+

obtained by applying $\mathrm{Th}$ to the map of vector bundles

$\begin{useimager} \[ \begin{tikzcd} \xi \ar[d] \ar[r] & \xi \boxplus 0 \ar[d]\\ X \ar[r, "\Delta"] & X \times X \end{tikzcd} \] \end{useimager}$

To see that the pullback of $\xi \boxplus 0$ along $\Delta$ is indeed $\xi$ , note that by definition, $\xi \boxplus 0$ is the pullback of $\xi$ along the projection onto the first factor.

Equivalently, if we view $\mathrm{Th}'(\xi )$ as $D(V) / S(V)$ , the space version of the map sends $v$ to $(v, \pi (v))$ where $\pi : D(V) \to X$ is the projection.

The final, correct form of the Thom isomorphism theorem then says

Theorem 1.7

Let $\xi : V \to X$ be an oriented vector bundle. Then there is a cohomology class $u: \mathrm{Th}(\xi ) \to H\mathbb {Z}$ such that the induced map

$\begin{useimager} \[ \begin{tikzcd} H\Z \wedge \Th(\xi) \ar[r, "H\Z \wedge \Delta"] & H\Z \wedge \Th(\xi) \wedge X_+ \ar[r, "H\Z \wedge u \wedge X_+"] & H\Z \wedge H\Z \wedge X_+ \ar[r, "\mu \wedge X_+"] & H\Z \wedge X_+ \end{tikzcd} \] \end{useimager}$

is an isomorphism of $H\mathbb {Z}$ -modules.