Does anyone know a good reference which proves Hopf degree theorem using Pontrjagin-Thom theorem, that is passing to the determination of framed bordism classes of 0-manifolds?

Many thanks!

Hopf degree theorem

Let $M$ be a closed connected manifold of dimension $n$

if $M$ is orientable then there exists an isomorphism $[M,S^n] \to \mathbb{Z}$

if $M$ is not orientable then there exists an isomorphism $[M,S^n] \to \mathbb{Z}_2$

Pontrjagin-Thom theorem

For any smooth compact $n$-manifold $M$ there is an isomorphism $\phi: [M,S^q] \to \Omega_{n-q;M}^{fr}$ which is the framed bordism group of $M$, where the forward map is the inverse image of a regular value and the inverse map is the Pontrjagin-Thom collapse map

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    $\begingroup$ In fact, this is done in Pontryagin's original paper! $\endgroup$ – Dylan Wilson Jul 12 '16 at 12:02
  • $\begingroup$ @DylanWilson: I see. Thanks for pointing that out. $\endgroup$ – PhysicsMath Jul 12 '16 at 22:53
  • $\begingroup$ @DylanWilson: could you do us a favor to post the link here, Dylan? $\endgroup$ – PhysicsMath Jul 12 '16 at 23:11
  • $\begingroup$ I don't know of a public link to that paper... I found it in a book by Novikov called "Topological Library I" where he has translated a bunch of classic papers into English. $\endgroup$ – Dylan Wilson Jul 13 '16 at 1:41

Let $M$ be a connected oriented closed manifold. We know $[M,S^n]$ is isomorphic to the group of cobordism classes of framed 0-dimensional submanifolds of $M$. A framed 0-dimensional submanifold of $M$ is a set of points in $M$ with an isomorphism $T_xM \cong \Bbb R^n$ at each. A single framed point is cobordant to the same point with a different framing if and only if the orientations agree; this is because $GL_n(\Bbb R)$ has two connected components and the manifold is oriented.

So there is a map $[M,S^n] \to \Bbb Z$ given by counting points (with orientation). I claim it's an isomorphism. It suffices to show that every framed 0-submanifold can be chosen so that it's either empty, or every point has the same orientation (positive or negative). For if there's a positively oriented point and a negatively oriented point, pick a path between them (that doesn't pass through of the other points) and frame the normal bundle of the path (which is trivial, as an interval is contractible); if you choose this to agree with the framing at the starting point, then by the assumption that the first point has a different orientation to the second point, the framing on the path agrees at both points. So if we have points whose orientations disagree, we can null-bord them both out of the picture, as desired.

This is slightly more subtle in the case of $n=1$, which I won't talk about (and in any case you can prove much more easily by hand).

In the nonorientable case, this falls apart in the first paragraph: every point is framed bordant to itself with the opposite orientation. Thus you can't just cancel points with the opposite orientation; you can cancel any pair of points, making the only well-defined invariant "number of points mod 2".

  • $\begingroup$ Many thanks Mike! But how about non-orientable $M$? Hopf degree theorem says $[M,S^n] \cong \mathbb{Z}_2$ i.e. we now have $\mathbb{Z}_2$ instead of $\mathbb{Z}$ degree. $\endgroup$ – PhysicsMath Jul 10 '16 at 0:56
  • $\begingroup$ @PhysicsMath An equivalent condition to non-orientability is that there is some loop $\gamma: [0,1] \to M$ with $\gamma(0)=\gamma(1)=x$ such that a framing of its normal bundle restricts to the opposite framings on $x$ at $0$ and $1$, aka the loop reverses orientation. Use this to prove that any pair of points (regardless of framing) is null-bordant, so that your only invariant is degree itself. $\endgroup$ – user98602 Jul 10 '16 at 0:58
  • $\begingroup$ I see! That is really concise and clear, Mike! Why don't you add this to your answer to make it more complete? By the way for the non-orientable case you said the only invariant is degree itself. What degree are you referring to? I thought that because you can null-bord any pair of points any framed 0-submanifold can either be empty or a single point and that gives $\mathbb{Z}_2$. Am I right? $\endgroup$ – PhysicsMath Jul 10 '16 at 1:18
  • $\begingroup$ @PhysicsMath Sorry, I wrote something nonsensical when I was thinking something else. $\endgroup$ – user98602 Jul 10 '16 at 1:38
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    $\begingroup$ @PhysicsMath It's the same argument. For instance, 2.46 is precisely the "point cancellation" idea above. The only difference is that I prefer to point out that the framing doesn't matter up to bordism for points, only the orientation; he does this implicitly in talking about how many path components $\mathcal B(M)$ has, but never explicitly. $\endgroup$ – user98602 Jul 12 '16 at 1:36

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