Complex oriented cohomology theoriesThom spaces

1 Thom spaces

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

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

        V \setminus X \ar[r] \ar[d] & V \ar[d]\\
        * \ar[r] & \Th'(\xi)

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

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

S(V)={vV:v=1},D(V)={vV:v1}. \begin{aligned} S(V) & = \{ v \in V: \| v\| = 1\} ,\\ D(V) & = \{ v \in V: \| v\| \leq 1\} . \end{aligned}


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

Example 1.2

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

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

  1. It is functorial along pullbacks — if f:XYf: X \to Y is a map and ξ:VY\xi : V \to Y is a vector bundle, then there is a natural map

    Th(f):Th(fξ)Th(ξ). \mathrm{Th}(f): \mathrm{Th}(f^* \xi ) \to \mathrm{Th}(\xi ).
  2. It is monoidal — if ξ:VX\xi : V \to X and ζ:WY\zeta : W \to Y are vector bundles and ξζ:VXX×Y\xi \boxplus \zeta : V \boxplus X \to X \times Y is the external direct sum. Then

    Th(ξζ)Th(ξ)Th(ζ). \mathrm{Th}'(\xi \boxplus \zeta ) \cong \mathrm{Th}'(\xi ) \wedge \mathrm{Th}'(\zeta ).

Corollary 1.3

Th(ξR)=ΣTh(ξ)\mathrm{Th}'(\xi \oplus \mathbb {R}) = \Sigma \mathrm{Th}'(\xi ).

ξR=ξ(triv:R)\xi \oplus \mathbb {R}= \xi \boxplus (\mathrm{triv}: \mathbb {R}\to *) and Th(triv)=S1\mathrm{Th}'(\mathrm{triv}) = S^1.

This leads to the following natural definition:

Definition 1.4

If ξ:VX\xi : V \to X is a rank dd vector bundle, then the Thom spectrum of ξ\xi is

Th(ξ)=ΣdΣTh(ξ). \mathrm{Th}(\xi ) = \Sigma ^{-d} \Sigma ^\infty \mathrm{Th}'(\xi ).
This definition is designed so that Th(ξR)=Th(ξ)\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:

Th(ξζ)Th(ξ)Th(ζ). \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 ξ:VX\xi : V \to X is an oriented vector bundle, then we have a natural isomorphism of HZH\mathbb {Z}-modules

HZTh(ξ)HZX+. 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 HZH\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 uH(Th(ξ))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

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

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

Definition 1.6

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

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

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

        \xi \ar[d] \ar[r] & \xi \boxplus 0 \ar[d]\\
        X \ar[r, "\Delta"] & X \times X

To see that the pullback of ξ0\xi \boxplus 0 along Δ\Delta is indeed ξ\xi , note that by definition, ξ0\xi \boxplus 0 is the pullback of ξ\xi along the projection onto the first factor.

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

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

Theorem 1.7

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

        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_+

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