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The Thom class and Thom isomorphism theorem for oriented vector bundles are proven ( at least to my knowledge) by induction on the open covers and some manipulation with Mayer-Vietoris sequences.

What is the "actual reason" behind the existence of Thom class? It seems strange that such an interesting class would exist just because some Mayer-Vietoris sequences routinely produce it.

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There are many fine answers already so I'll just add this: if you want understand the "actual reason" for the existence of something (like the Thom class), you probably ought to try to understand the way it can fail to exist. In this case, why do non-orientable bundles fail to have a Thom class (in ordinary cohomology with integer coefficients)? – Charles Rezk Sep 22 '12 at 15:43
up vote 37 down vote accepted

It is easy to understand the existence of a Thom class by considering cellular cohomology. Let the given vector bundle be $E\to B$ with fibers of dimension $n$. One can assume without significant loss of generality that $B$ is a CW complex with a single 0-cell. The Thom space $T(E)$ is the quotient $D(E)/S(E)$ of the unit disk bundle of $E$ by the unit sphere bundle. One can give $T(E)$ a CW structure with $S(E)/S(E)$ as the only 0-cell and with an $(n+k)$-cell for each $k$-cell of $B$. These cells in $T(E)$ arise from pulling back the bundle $D(E)\to B$ via characteristic maps $D^k\to B$ for the $k$-cells of $B$. These pullback are products since $D^k$ is contractible.

In particular, $T(E)$ has a single $n$-cell and an $(n+1)$-cell for each 1-cell of $B$. There are no cells in $T(E)$ between dimension $0$ and $n$. The cellular boundary of an $(n+1)$-cell is 0 if $E$ is orientable over the corresponding 1-cell of $B$, and it is twice the $n$-cell in the opposite case. Thus $H^n(T(E);{\mathbb Z})$ is $\mathbb Z$ if $E$ is orientable and $0$ if $E$ is non-orientable. In the orientable case a generator of $H^n(T(E);{\mathbb Z})$ restricts to a generator of $H^n(S^n;{\mathbb Z})$ in the "fiber" $S^n$ of $T(E)$ over the 0-cell of $B$, hence the same is true for all the "fibers" $S^n$ and so one has a Thom class.

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Thank you Allen, your answer clarifies things, especially for me, a non-expert in topology. I think following your simple $CW$-reasoning it should be more or less easy to see why cupping with Thom class the cohomology of $B$ one has Thom isomorphism. – Axel Sep 23 '12 at 8:48

One not-very technical way to think of the Thom Isomorphism Theorem is that if you have a vector bundle, $p : E \to B$, if you remove the $0$-section $Z$ of the vector bundle from the Thom space $Th(p)$, you get a contractible space. So given a homology class in $H_* Th(p)$ the obstruction to trivializing it can be thought of as its intersection with $Z$. If there's no intersection, you're in the contractible space $Th(p) \setminus Z$. So the intersection of a homology class with $Z$ is tautologically the thing that keeps track of the homology class itself.

That's how I like to think of the Thom Isomorphism Theorem. So why is there a Thom class? Because you can intersect with $Z$. In cohomology this is cupping with the Thom class since that's what intersections translate to in cohomology.

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You are thinking in terms of ordinary cohomology, where Mayer-Vietoris patches together the always present local orientation to produce a global one when you have it. It is more advanced, but maybe more illuminating, to note that the definition in general is intrinsically global. An $n$-plane bundle $p$ over a space $B$ has an associated sphere bundle $Sph(p)$ (by fiberwise one point compactification) with based fibers and thus a section. The quotient $Sph/B$ is the Thom space $T$ of $p$. For a multiplicative cohomology theory $E$, a Thom class $\mu$ is an element of $\tilde{E}^n(T)$ whose restriction to $\tilde{E}^n(S^n_b)\cong \tilde{E}^0(S^0)$ is a unit in this ring for any $b\in B$, where $S^n_b$ is the fiber over $b$ in $Sph(p)$. This definition is admitttedly mysterious. It suffices to give a Thom isomorphism and it is important geometrically, but the real explanation is more advanced and still not very well known. One should think of $E^*$ as represented by a ring spectrum $E$. Bundle theory naturally concerns spaces over $B$, or parametrized spaces. One can make sense of parametrized spectra over $B$, and one can even take the smash product of a parametrized space and a spectrum to obtain a parametrized spectrum. Thus one can make sense of $Sph(p)\wedge E$ as a spectrum over $B$. Of course, there is also a trivial spherical bundle $B\times S^n$ over $B$. It turns out that a Thom class as I defined it cohomologically is the same thing as a trivialization: an equivalence of parametrized spectra between $Sph(p)\wedge E$ and $(B\times S^n)\wedge E$. That is the geometric meaning. This is proven in the book Parametrized Homotopy Theory, by Sigurdsson and myself (available on my website).

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Tom, your comment is incomplete and needs editing. The theorem that is unstated needs its hypotheses (compatibity on intersections) as well as its statement. But of course the point that local implies global fails for generalized cohomology is part of what I had in mind. (While the question implicitly refers to ordinary cohomology, that is not explicit, so it seemed not unreasonable to give a general answer). – Peter May Sep 22 '12 at 17:11
Yes, it's hard that comments can't be edited. I carelessly lost one set of words in dividing the comment, and I also carelessly forgot to mention the local triviality hypothesis. – Tom Goodwillie Sep 23 '12 at 15:45
I've deleted it now. Of course I like your advanced viewpoint; I just couldn't see it as an answer to the question. Your answer could be fleshed out to make the point that the (reduced) ordinary cohomology of $S^n$ vanishes in degrees less than $n$ (i.e. that the coefficient groups of ordinary cohomology vanish in positive degrees), which is what makes the existence of a Thom class follow from the existence of an orientation. The same point could be made in a less advanced way using a spectral sequence rather than parametrized spectra. – Tom Goodwillie Sep 23 '12 at 15:47

The idea behind the Thom isomorphism $\beta:H^iX \rightarrow H^{n+i}(DE,SE)$ is implicit in the formula $$\int_{\sigma_{n+i}} \beta(\alpha_i) = \int_{X\cap \sigma_{n+i}} \alpha_i$$ Here $\sigma_{n+i}$ is a singular simplex in $DE$ and we have written integration for the evaluation of a cochain on a sum of simplices. Also $X\subset DE$ is identified with the zero-section.

The problem with this formula is that it doesn't make sense in full generality: after all $X\cap\sigma_{n+i}$ will not in general be a simplex again. And even if it is, it might be a simplex in many different ways (different parametrizations), so some choices must be made. These problems can be overcome and this is the "miracle" of the Thom isomorphism.

Note that the right hand side also requires an "orientation" of $X\cap\sigma_{n+i}$. This is why you also require an orientation on $E$.

For the Thom class $\tau = \beta(1)$ itself this gives the characterization $$\langle \tau, \sigma_n\rangle = \sharp ( X \cap \sigma_n )$$ where the intersection points are counted with appropriate signs. (In $DE$ a generic $n$-simplex has a zero-dimensional intersection with the zero section.)

You might find it helpful to learn something about Thom classes in other (generalized) cohomology theories: in de Rham cohomology and K-theory there are pretty explicit representatives for the respective Thom classes. And nothing beats the elegance of Thom classes in cobordism theories, where you've got a "tautological" Thom class.

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Even the case of an oriented vector bundle over a point, which is where the story begins, is nontrivial. In this case the Thom isomorphism is the Poincare duality for the cohomology with compact supports on an oriented vector space. Ultimately, the Thom isomorphism theorem is a special form of the Poincare-Verdier duality. The fact that the Mayer-Vietoris technique is used in the proof is an indication that the Thom isomorphism deals with the cohomologies of some sheaves.

If the base of the vector bundle is compact and oriented, then the Thom isomorphism is equivalent to the Poincare-Lefschetz duality for an oriented manifold with boundary namely, the unit disk bundle determined by the vector bundle.

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Thom class gives an orientation covector in every fiber $F\cong\mathbb R^n$ (of an oriented vector bundle) which can thought of a generator in $H^n(F-0)$ . Using local trivializations such covectors are defined locally. One needs to prove that these covectors glue to a cohomology class on the total space (with the zero section deleted), and this is where Mayer-Vietoris becomes relevant. How else would you glue? Read the exposition in Milnor-Stasheff or Bott-Tu.

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My intuition (maybe misleading?) of the Thom isomorphism is this: just as a vector bundle $\xi:X\to B$ is a twist of the trivial bundle, its Thom space $T\xi$ is a twist of the (rank $\xi$)-fold suspension of $B$ (Thom space of the trivial bundle is just the iterated suspension).

Now for a multiplicative cohomology theory $E$, it seems that this twist manifests itself in the fact that $\tilde E^*(T\xi)$ is a rank 1 projective module over $E^*(B)$, i. e. a twist of the free rank 1 module. And picking a $E^*$-orientation of $\xi$ is more or less the same as picking a generator (necessarily of degree rank $\xi$) of this module; in particular, such thing exists iff this module is free, and then the Thom isomorphism is clear - it is just dimension shift by degree of the generator.

Thus one may say that a bundle is $E$-orientable iff $E$ "is not confused by the twist of the iterated suspension introduced by the twist of the trivial bundle caused by $\xi$".

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A downvote indicates that something's wrong with such intuition. I would be very grateful for an explanation, it might be very important for my work - few things are more dangerous in mathematical work than wrong intuitions! – მამუკა ჯიბლაძე May 18 '14 at 4:18
I see your answer as more of a formalistic encoding of the Thom isomorphism (once you have it) rather than a justification for why it exists. – Ryan Budney May 18 '14 at 4:28
@RyanBudney I see thank you. So at least it's harmless... – მამუკა ჯიბლაძე May 18 '14 at 5:12

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