I am trying to understand the Pontryagin-Thom theorem; especially how the Thom space comes into play. Just to bring everyone on the same page: I am specifically talking about the construction of an isomorphism $\Omega_n^k \cong \pi_{n + k}(MO(k))$, where $\Omega_n^k$ is the set of bordism classes of $n$-manifolds embedding in $\mathbb{R}^{n+k}$ and $MO(k) = \mathrm{Th}(\xi)$ is the Thom space of the universal $k$-plane bundle $\xi\colon EO(k) \to BO(k)$.

I find the construction of the Thom space rather peculiar, so I was looking around for intuition behind its role in this theorem. I found these notes, which, on page 4, hint at an approach to introduce the Thom space which I find quite interesting:

We want to classify (bordism classes of) $n$-manifolds which embed in $\mathrm{R}^{n+k}$ by homotopy classes of maps from $S^{n+k}$. So given such a manifold $M$, we are tasked with finding such a map that carries some information of $M$. Now we know that $S^{n+k}$ is the one point compactification of $\mathbb{R}^{n+k}$ and we know that $M$ embeds into this space openly, when we consider a tubular neighbourhood.

But given an open embedding $U \hookrightarrow S$, we can contravariantly construct a map $S^{+} \to U^{+}$ between the one point compactifications, so in our case, the open embedding $i\colon N \to \mathbb{R}^{n+k}$ of the normal bundle $\nu\colon N \to M$ gives us a map $S^{n+k} \to N^{+}$ which carries quite a lot of information about $M$.

The text assumes $M$ to be compact in this context and then goes on to present the Thom space as a generalisation of this idea that also works for non compact base spaces.

The question, at last

Does the above idea not work with non compact $M$? If not: why?

If it does, where else in the remainder of the proof of the above theorem does a simple one point compactification stop to "work"? And what is it that then makes the Thom space the "correct" generalisation?

I do have one idea, but I'm absolutely not sure whether this is anywhere near the right answer: One point compactification is not functorial, right? And the next step in our construction would be to extend the map constructed above by the bundle map that covers the classifying map $f\colon M \to BO(k)$ of $\nu$, but we would have to apply one point compactification to it as well. As far as I know, this only works for proper maps, and we don't know $f$ to be proper?

According to the above notes, the Thom construction is functorial, but only on bundle maps which are injective on each fiber; do we have that in our situation? And is this really the crux of the matter?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.