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The Pontryagin-Thom construction gives an isomorphism from the stable homotopy groups of spheres and framed cobordism groups. It seems to be well-established that for dimension 1 (see this question), the circle with the Lie group framing on the tangent bundle represents the Hopf map $\eta$. My first question is the following:

  1. How does one prove that this is the case?

Here's an attempt: since the $\pi_1^s\simeq \mathbb{Z}/2,$ one only needs to show that this manifold cannot be the boundary of a framed surface. Inspired by this post, if this was the case, then this surface would have a vector field with one isolated zero of index 1. By Poincaré-Hopf, it follows that its Euler characteristic is 1, but oriented surfaces have even Euler characteristic.

However, this proof is not really what I am looking for. I would like to know why the Pontryagin Thom map applied to the Hopf fibration gives this framing. Technically, this construction gives a framing on the normal bundle. According to this question, the Hopf map represents the unknot on $S^3$ with the framing that "twists once", which brings to a vaguer question:

  1. Why does the stable normal framing that "twists once" correspond to the Lie group tangential framing?

Another problem that the above proof has is that it does not generalize to prove that the quaternionic Hopf map corresponds to the 3-sphere with its Lie group framing. I believe that somehow this question has to do with the fact that the Hopf map (resp. quaternionic Hopf map) is a $U(1)\simeq S^1$ (resp. $SU(2)\simeq S^3$) principal bundle.

Any insights would be appreciated!

Thank you very much!

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2 Answers 2

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I wrote out careful proofs of all of this and more in my note "Homotopy groups of spheres and low-dimensional topology", available here.

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  • $\begingroup$ Hi Andy, did these notes ever get published somewhere? $\endgroup$ Commented Apr 13, 2023 at 20:39
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    $\begingroup$ @DaveBenson: No, I never did publish them. I've thought about sending them somewhere, but somehow never got around to it. $\endgroup$ Commented Apr 13, 2023 at 20:59
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Here is a direct answer (computation) to your question.

Let E be a bundle over a manifold M. The first thing to notice is that framing of E is a choice of basis (for each fiber), there is no canonical basis for E to compare to, so if you have only one framing, you can't ask "what framing I have?".
If you have two framing of E, you get two basis for E, and two basis give you a change of basis matrix, for each fiber. Now you can ask "do these two framings are equivalent?". I our case (where $M=S^1$) we just have a $S^1$ family of matrices in $SO$, i.e. an element in $\pi_1(SO)$ and the two framing are equivalent if this element is trivial.

The second thing to notice is that Lie framing, is actually a parallelization, and the Hopf framing, is a normal framing / stable framing. To compare there we need to "stabilize" the Lie framing, and we get two framings for the bundle.
Lets see an example, by comparing the Lie framing with the Pontryagin Thom of the map $f:\mathbb{R}^2\rightarrow \mathbb{R}$ which is $f(x,y)=x^2+y^2$. The framing of the preimage $S^1=f^{-1}(1)$ coming from the differential of $f$ which trivialize the normal bundle by taking the "out" normal, for example and using the embedding of $S^1$ in $\mathbb{R}^2$ to trivialize the direct sum $TS^1\oplus \nu S^1$, so we actually get a farming of the bundle $TS^1\oplus \mathbb{R}$ (which explains the name "stable framing"). Now, we compare the two framings, and we are doing so by taking $\lambda \in S^1$ and looking at $$\mathbb{R}^2\simeq \mathbb{R}\oplus \mathbb{R} \simeq T_eS^1 \oplus \mathbb{R} \simeq T_{\lambda}S^1 \oplus \mathbb{R} \simeq T_{\lambda}S^1 \oplus \nu_{\lambda} S^1 \overset{!}{\simeq} T_{\lambda}\mathbb{R}^2\simeq \mathbb{R}^2 $$ Where the first three isomorphism are the (stabilization of the) Lie framing, and the last three are the framing that coming from $f$.
Notice to the "!" isomorphism; the composition, until "!", take the basis $b,b^{\prime}$ of $\mathbb{R}\oplus\mathbb{R}$ and send it to the tangent vector (The Lie framing) and the normal vector (taking "out" normal using $df^{-1}$) but the "!" isomorphism is the transformation matrix from the "tangent and normal" basis to the standard basis of $\mathbb{R}^2$ which is the rotation matrix, by $\lambda$ radians (where we think of $S^1$ as $(\cos \lambda,\sin \lambda)$). Hence, the loop in $SO(2)$ is the loop that start with rotation by 0 and ends with rotation by $2\pi$ which is non-trivial element in $\pi_1(SO(2))$, and we conclude that the the Lie framing isn't the Pontryagin Thom framing of the map $f$.

Now, to the Hopf map.
The Hopf map is given by the map $\eta:\mathbb{R}^4\rightarrow \mathbb{R}^3$ $$(2(xz+wy),2(yz-wx),1-2(x^2+y^2)$$ The differential given by $$ 2\left(\begin{array}{cccc} z & w & x & y\\ -w & z & y & -x\\ -x & -y & 0 & 0 \end{array}\right)$$ We take the preimage of $(0,0,-1)$ which is $$\eta^{-1}\left(0,0-1\right)=\left\{ \left(x,y,0,0\right)|x^{2}+y^{2}=1\right\}$$ and get that the differential in the points of the form $(x,y,0,0)$ is $$2\left(\begin{array}{cccc} 0 & 0 & x & y\\ 0 & 0 & y & -x\\ -x & -y & 0 & 0 \end{array}\right)$$ This induces the isomorphism $\mathbb{R}^{3}\simeq\nu_{\left(x,y\right)}S^{1}$ by pulling a basis of $\mathbb{R}^{3}$ with $d_{\left(x,y\right)}\eta^{-1}$. We see that $$ \left(\begin{array}{cccc} 0 & 0 & x & y\\ 0 & 0 & y & -x\\ -x & -y & 0 & 0 \end{array}\right)\left(\begin{array}{c} x\\ y\\ 0\\ 0 \end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ -1 \end{array}\right) $$ $$ \left(\begin{array}{cccc} 0 & 0 & x & y\\ 0 & 0 & y & -x\\ -x & -y & 0 & 0 \end{array}\right)\left(\begin{array}{c} 0\\ 0\\ x\\ y \end{array}\right)=\left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right) $$ $$ \left(\begin{array}{cccc} 0 & 0 & x & y\\ 0 & 0 & y & -x\\ -x & -y & 0 & 0 \end{array}\right)\left(\begin{array}{c} 0\\ 0\\ -y\\ x \end{array}\right)=\left(\begin{array}{c} 0\\ -1\\ 0 \end{array}\right) $$ so the basis for $\nu_{\left(x,y\right)}S^{1}$ is given (in the standard basis) by $$ \left(\begin{array}{ccc} 0 & 0 & x\\ 0 & 0 & y\\ x & -y & 0\\ y & x & 0 \end{array}\right) $$ and the tangent is given by $\left(-y,x,0,0\right)$. Now, like we did for the map $f$ framing, we need to calculation the transformation matrix which "measure" the difference between the Lie and he Hopf. As we saw, it is given by "taking the tangent, taking a normal to the basis using the differential of $\eta$, and write that in the standard basis (the "!" isomorphism)" So we just need to combing all the above, and get that that for a point $\left(x,y,0,0\right)\in S^{1}\hookrightarrow\mathbb{R}^{4}$ the transformation matrix in $SO\left(4\right) $ is given by $$ \left(\begin{array}{cccc} 0 & 0 & x & -y\\ 0 & 0 & y & x\\ x & -y & 0 & 0\\ y & x & 0 & 0 \end{array}\right) $$ and the element in $\pi_{1}(SO(4))$ is multiplication of two “simple” rotations matrices (the one that we saw in the map $f$). Using Eckmann–Hilton argument we can find that composition of two loops in $\pi_{1}(SO(n))$ is like concatenation so the element in $\pi_1(SO(4))$ is sum of two non trivial elements. Use the fact that $\pi_1(SO(4))=\mathbb{Z}/2$ and we get that the difference between the Lie framing, and the Hopf framing is a trivial element in $\pi_{1}(SO(4))$ so these two framings are equivalent!.

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  • $\begingroup$ Thank you for the explicit anwer, that helps a lot. $\endgroup$ Commented Apr 17, 2023 at 14:28

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