Editing Thom space (section)

==The Thom isomorphism==

The significance of this construction begins with the following result, which belongs to the subject of [[cohomology]] of [[fiber bundle]]s. (We have stated the result in terms of <math>\Z_2</math> [[coefficients]] to avoid complications arising from [[orientability]]; see also [[Orientation of a vector bundle#Thom space]].)

Let <math>p: E\to B</math> be a real vector bundle of rank ''n''. Then there is an isomorphism called a '''Thom isomorphism'''
:<math>\Phi : H^k(B; \Z_2) \to \widetilde{H}^{k+n}(T(E); \Z_2),</math>
for all ''k'' greater than or equal to 0, where the [[Left-hand side and right-hand side of an equation|right hand side]] is [[reduced cohomology]].

This theorem was formulated and proved by [[René Thom]] in his famous 1952 thesis.

We can interpret the theorem as a global generalization of the suspension isomorphism on local trivializations, because the Thom space of a trivial bundle on ''B'' of rank ''k'' is isomorphic to the ''k''th suspension of <math>B_+</math>, ''B'' with a disjoint point added (cf. [[#Construction of the Thom space]].) This can be more easily seen in the formulation of the theorem that does not make reference to Thom space:

{{math_theorem|name=Thom isomorphism|
Let <math>\Lambda</math> be a ring and <math>p: E\to B</math> be an [[oriented bundle|oriented]] real vector bundle of rank ''n''. Then there exists a class
:<math>u \in H^n(E, E \setminus B; \Lambda),</math>
where ''B'' is embedded into ''E'' as a zero section, such that for any fiber ''F'' the restriction of ''u''
:<math>u|_{(F, F \setminus 0)} \in H^n(F, F \setminus 0; \Lambda)</math>
is the class induced by the orientation of ''F''. Moreover,
:<math>\begin{cases} H^k(E; \Lambda) \to H^{k+n}(E, E \setminus B; \Lambda) \\ x \longmapsto x \smile u \end{cases}</math>
is an isomorphism.
}}

In concise terms, the last part of the theorem says that ''u'' freely generates <math>H^*(E, E \setminus B; \Lambda)</math> as a right <math>H^*(E; \Lambda)</math>-module. The class ''u'' is usually called the '''Thom class''' of ''E''. Since the pullback <math>p^*: H^*(B; \Lambda) \to H^*(E; \Lambda)</math> is a [[ring isomorphism]], <math>\Phi</math> is given by the equation:

:<math>\Phi(b) = p^*(b) \smile u.</math>

In particular, the Thom isomorphism sends the [[identity (mathematics)|identity]] element of <math>H^*(B)</math> to ''u''. Note: for this formula to make sense, ''u'' is treated as an element of (we drop the ring <math>\Lambda</math>)

:<math>\tilde{H}^n(T(E)) = H^n(\operatorname{Sph}(E), B) \simeq H^n(E, E \setminus B).</math><ref>Proof of the isomorphism. We can embed ''B'' into <math>\operatorname{Sph}(E)</math> either as the zero section; i.e., a section at zero vector or as the infinity section; i.e., a section at infinity vector (topologically the difference is immaterial.) Using two ways of embedding we have the triple:
:<math>(\operatorname{Sph}(E), \operatorname{Sph}(E) \setminus B, B)</math>.
Clearly, <math>\operatorname{Sph}(E) \setminus B</math> deformation-retracts to ''B''. Taking the long exact sequence of this triple, we then see:

:<math>H^n(Sph(E), B) \simeq H^n(\operatorname{Sph}(E), \operatorname{Sph}(E) \setminus B),</math>

the latter of which is isomorphic to:

:<math>H^n(E, E \setminus B)</math>

by excision.</ref><!-- I can't understand this:
Note: for this formula to make sense, ''u'' is treated as an element of <math>H^k(D(E),\operatorname{Sph}(E);\Z_2)\cong \tilde H^k(T(E);\Z_2)</math>, where <math>D(E)</math> is the associated disk bundle, so we have a cup product

:<math>H^i(D(E);\Z_2)\otimes H^k(D(E),\operatorname{Sph}(E);\Z_2)\to H^{i+k}(D(E),\operatorname{Sph}(E);\Z_2)\cong \tilde H^k(T(E);\Z_2)</math>.-->

The standard reference for the Thom isomorphism is the book by Bott and Tu.