# Thom isomorphism theorem

Let $\xi\to X$ be a $d$-dimensional vector bundle over a topological space $X$, and let $h^{*}$ be a multiplicative generalized cohomology theory, such as ordinary cohomology. Let $\tau\in h^{d}(D(\xi),S(\xi))$ be a Thom class for $\xi$, where $D(\xi)$ and $S(\xi)$ are the associated disk and sphere bundles of $\xi$.

Since $h^{*}$ is a multiplicative theory, there is a generalized cup product map

 $h^{*}(D(\xi))\otimes_{h^{*}}h^{*}(D(\xi),S(\xi))\to h^{*}(D(\xi),S(\xi)),$

where the tensor product is over the coefficient ring $h^{*}(\mathrm{pt})$ of the theory. Using the isomorphism $p^{*}:h^{*}(X)\cong h^{*}(D(\xi))$ induced by the homotopy equivalence $p:D(\xi)\to X$, we obtain a homomorphism

 $T:h^{n}(X)\to h^{n+d}(D(\xi),S(\xi))\cong\tilde{h}^{n+d}(X^{\xi})$

taking $\alpha$ to $p^{*}(\alpha)\cdot\tau$. Here $X^{\xi}$ stands for the Thom space $D(\xi)/S(\xi)$ of $\xi$.

Thom isomorphism theorem$T$ is an isomorphism $h^{*}(X)\cong\tilde{h}^{*+d}(X^{\xi})$ of graded modules over $h^{*}(\mathrm{pt})$.

###### Remark 1

When $\xi$ is a trivial bundle of dimension $1$, this generalizes the suspension isomorphism. In fact, a typical proof of this theorem for compact $X$ proceeds by induction over the number of open sets in a trivialization of $\xi$, using the suspension isomorphism as the base case and the Mayer-Vietoris sequence to carry out the inductive step.

###### Remark 2

There is also a homology Thom isomorphism $\tilde{h}_{*+d}(X^{\xi})\cong h_{*}(X)$, in which the map is given by cap product with the Thom class rather than cup product.

Title Thom isomorphism theorem ThomIsomorphismTheorem 2013-03-22 15:40:52 2013-03-22 15:40:52 antonio (1116) antonio (1116) 6 antonio (1116) Theorem msc 55-00