Thom class

Let $h^{*}$ be a generalized cohomology theory (for example, let $h^{*}=H^{*}$, singular cohomology with integer coefficients). Let $\xi \to X$ be a vector bundle of dimension $d$ over a topological space $X$. Assume for convenience that $\xi$ has a Riemannian metric, so that we may speak of its associated sphere and disk bundles, $S(\xi )$ and $D(\xi )$ respectively.

Let $x\in X$, and consider the fibers $S({\xi }_x)$ and $D({\xi }_x)$. Since $D({\xi }_x)/S({\xi }_x)$ is homeomorphic to the $d$-sphere, the Eilenberg-Steenrod axioms for $h^{*}$ imply that $h^{*+d}(D({\xi }_x),S({\xi }_x))$ is isomorphic to the coefficient group $h^{*}(\mathrm{pt})$ of $h^{*}$. In fact, $h^{*}(D({\xi }_x),S({\xi }_x))$ is a free module of rank one over the ring $h^{*}(\mathrm{pt})$.

Definition 1

An element $\tau \in h^{*}(D(\xi ),S(\xi ))$ is said to be a Thom class for $\xi$ if, for every $x\in X$, the restriction of $\tau$ to $h^{*}(D({\xi }_x),S({\xi }_x))$ is an $h^{*}(\mathrm{pt})$-module generator.

Note that $\tau$ lies necessarily in $h^d(D(\xi ),S(\xi ))$.

Definition 2

If a Thom class for $\xi$ exists, $\xi$ is said to be orientable with respect to the cohomology theory $h^{*}$.

Remark 1

Notice that we may consider $\tau$ as an element of the reduced $h^{*}$-cohomology group ${\tilde{h}}^{*}(X^{\xi })$, where $X^{\xi }$ is the Thom space $D(\xi )/S(\xi )$ of $\xi$. As is the case in the definition of the Thom space, the Thom class may be defined without reference to associated disk and sphere bundles, and hence to a Riemannian metric on $\xi$. For example, the pair $(\xi ,\xi -X)$ (where $X$ is included in $\xi$ as the zero section) is homotopy equivalent to $(D(\xi ),S(\xi ))$.

Remark 2

If $h^{*}$ is singular cohomology with integer coefficients, then $\xi$ has a Thom class if and only if it is an orientable vector bundle in the ordinary sense, and the choices of Thom class are in one-to-one correspondence with the orientations.