The Thom-Space of a vector bundle $\pi: V \to M$ is defined as the quotient of the total space of the associated disk bundle $D(V) \to M$ by the total space of the associated sphere bundle $S(V)$, so $$ \text{Th}(V) = D(V)/S(V) $$ what is the original motivation behind this construction?

  • $\begingroup$ I don't know whether this is the original motivation, but I'm sure it was noticed pretty early on that Thom spaces are related to cobordism groups, which people care about. $\endgroup$ – JHF May 5 '17 at 14:44
  • $\begingroup$ @JHF how are they related to cobordism groups? $\endgroup$ – 54321user May 5 '17 at 15:29

I brought up the connection between the homotopy groups of Thom spaces and cobordism groups in the comments, so I'll expand on it in this answer. What I'm about to describe is the Pontryagin-Thom construction, which is covered in many other places, but here is a quick summary.

Suppose I have a vector bundle $V$ of rank $n$ over $M$, and suppose I have a map $f: S^{n+k} \to \operatorname{Th}(V)$ that is transverse to the zero section $Z \subset \operatorname{Th}(V)$. Then we consider $X := f^{-1}(Z)$, which is a $k$-dimensional manifold that is equipped with

  • an embedding $X \hookrightarrow \mathbb{R}^{n+k} \subset S^{n+k}$,
  • a map $f|_X: X \to M$, and
  • a framing of the normal bundle $Df: \nu \to f^* V$.

Moreover, varying $f$ within its homotopy class yields manifolds $X'$ that are cobordant to $X$. In fact, we have

Theorem (Pontryagin-Thom). There is a bijection between $\pi_{n+k} \operatorname{Th}(V)$ and $k$-dimensional manifolds equipped with the above data, up to cobordism.

To actually get the cobordism group $\mathfrak{N}_*$, we have to modify the construction so as to "forget" the extra data. This can be done by

  1. Passing to stable homotopy groups.
  2. Taking $V$ to be the universal rank $n$ bundle over $M = BO(n)$. Denote the Thom space in this case by $MO(n)$.

Once we do this, we arrive at the isomorphism $\mathfrak{N}_* \cong \operatorname{colim}_k \pi_{n+k} MO(k)$.

This is the beginning of many different stories in geometric topology, stable homotopy theory, surgery theory, ....


The Thom space is some kind of "twisted suspension". If $\pi$ is a trivial vector bundle, say $V = M \times \mathbb{R}^n$, then $\operatorname{Th}(V)$ is the $n$th suspension of $M \cup \{*\}$ (the space $M$ with a disjoint point added). Locally $\operatorname{Th}(V)$ "looks like" the $n$th suspension of $n$ (because locally the bundle is trivial), but globally the whole thing is twisted.

We know that for suspensions we have a suspension isomorphism, $H^k(M) \cong \tilde{H}^{k+n}(S^nM)$. Well, there is an analogous isomorphism for Thom spaces, called the Thom isomorphism: $$H^k(M) \cong \tilde{H}^{k+n}(\operatorname{Th}(V)).$$

This can be rephrased without any mention of the Thom space (though you need it for the proof AFAIK). There exists some class $u \in H^n(V, V \setminus M)$ (where $M$ is embedded in $V$ as the zero section), called the Thom class, such that for all $k$ the following map is an isomorphism: $$H^k(E) \to H^{k+n}(E, E \setminus M), \; x \mapsto x \smile u.$$

For more information, see the Wikipedia article.

  • $\begingroup$ This theorem is cool and all, but I can't help but saying: "so what?" (in the pedagogical, not sarcastic sense) $\endgroup$ – 54321user May 5 '17 at 16:30

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