# 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

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

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

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

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

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

This leads to the following natural definition:

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

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

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

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

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

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

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

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