Thom isomorphism theorem
Let be a -dimensional vector bundle over a topological space , and let be a multiplicative generalized cohomology theory, such as ordinary cohomology. Let be a Thom class for , where and are the associated disk and sphere bundles of .
Since is a multiplicative theory, there is a generalized cup product map
where the tensor product is over the coefficient ring of the theory. Using the isomorphism induced by the homotopy equivalence , we obtain a homomorphism
taking to . Here stands for the Thom space of .
Thom isomorphism theorem is an isomorphism of graded modules over .
Remark 1
When is a trivial bundle of dimension , this generalizes the suspension isomorphism. In fact, a typical proof of this theorem for compact proceeds by induction over the number of open sets in a trivialization of , 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 , in which the map is given by cap product with the Thom class rather than cup product.
Title | Thom isomorphism theorem |
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Canonical name | ThomIsomorphismTheorem |
Date of creation | 2013-03-22 15:40:52 |
Last modified on | 2013-03-22 15:40:52 |
Owner | antonio (1116) |
Last modified by | antonio (1116) |
Numerical id | 6 |
Author | antonio (1116) |
Entry type | Theorem |
Classification | msc 55-00 |