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In the construction of exotic 7-spheres, Milnor used the fact the first Pontrjagin class $p_1(\xi_{ij})$ is linear in $i,j$. He claims that this is clear but I don't feel that. I tried to find some other notes about it. Then I found some more rigorous explanations, and they usually go as follows.

Recall that $\xi_{ij}$ is a rank 4 oriented vector bundle over $S^4$ obtained from the $[f_{ij}]\in \pi_3(SO(4)) \cong \mathbb Z \oplus \mathbb Z$, clutching function $f_{ij}(u)v= u^i \cdot v \cdot u^j$, where $\cdot$ stands for the quaternion multiplication.

The universal bundle $ESO(4) \rightarrow BSO(4)$ gives rise to an isomorphism $\pi_4(BSO(4)) \cong \pi_3(SO(4))$ (the connecting homomorphism $\delta$ of the homotopy l.e.s).

On the other hand, it can be verified that the map $\Theta: \pi_4(BSO(4)) \rightarrow H^4(S^4)$ defined by $[f] \mapsto p_1(f^* ESO(4))$ is actually a homomorphism. Therefore, by composing with the connecting map $\delta$, we obtain a homomorphism $$ \mathbb Z \oplus \mathbb Z \cong \pi_3(SO(4)) \rightarrow \pi_4(BSO(4)) \rightarrow H^4(S^4)$$

that sends $(i,j)$ to $p_1(\xi_{ij})$. It seems that we are done from here, but in this proof, we implicitly assumed two facts.

First, we used that the bundle obtained from the clutching function in $\pi_3(SO(4))$ and the bundle obtained from the pull-back of the universal bundle by the corresponding map of $\pi_4(BSO(4))$ (the preimage of the clutching function via the connecting homomorphism $\delta$) are the same, i.e. the following diagram

$$\require{AMScd} \begin{CD} \pi_4(BSO(4)) @>\delta>> \pi_3(SO(4)\\ @.{_{\rlap{f^*}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{4em}{0em}]{}}}} @VVV\\ @. \mathrm{Vect}^{+}_{4}(S^4) \end{CD}$$

commutes.

Second, we also assumed that $[f_{i+i',j+j'}] = [f_{ij}] + [f_{i'j'}] \in \pi_3(SO(4))$.

Can someone tell me why these are true?

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2 Answers 2

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  1. This is a general fact. If $G$ is a topological group and $f\colon S^{n+1}\rightarrow BG$ classifies a principal $G$-bundle $p\colon E\rightarrow S^{n+1}$, then the naturality of the fibration LES tells you that the image of $[f]$ under $\pi_{n+1}(BG)\stackrel{\sim}{\rightarrow}\pi_n(G)$ takes $[f]$ to the image of $\mathrm{id}_{S^{n+1}}$ under the boundary map $\partial\colon\pi_{n+1}(S^{n+1})\rightarrow\pi_n(G)$ for the fibration $p$. Now, choose trivializations $\varphi_{\pm}\colon E\vert_{D^n_{\pm}}\rightarrow D^n_{\pm}\times G$ over the hemispheres and let $\psi\colon S^n\rightarrow G$ be the induced clutching map. Then, $S^n\times [-1,1]\rightarrow E,\,(x,t)\mapsto\varphi_-^{-1}(\pi(x,t),e)$ for $t\le0$ and $\varphi_+^{-1}(\pi(x,t),\psi(x))$ for $t\ge0$ define a lift of the quotient map $\pi\colon S^n\times[-1,1]\rightarrow S^{n+1}$ that is constant on $S^n\times\{-1\}$, so its restriction to $S^n\times\{1\}$ defines $\partial[\mathrm{id}_{S^{n+1}}]$, but this is the clutching map $\psi$ by construction (I'm taking a cavalier attitude towards basepoints, forgive me).
  2. The definition implies that $f_{i+i^{\prime},j+j^{\prime}}=f_{i^{\prime},j^{\prime}}\cdot f_{ij}$ (this means pointwise matrix multiplication), but the Eckmann-Hilton argument tells you that the pointwise group multiplication of $SO(4)$ does in fact induce the usual addition on $\pi_3(SO(4))$.
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You could argue that the bundles $E_{[f]}, E_{[g]} \to S^4$ defined by $[f]$ and $[g] \in \pi_3(SO(4))$ are related to $E_{[f]+[g]}$ as follows. Consider the map $\pi : S^4 \to S^4 \vee S^4$ that crushes an equator to a point. Define a bundle $E' \to S^4 \vee S^4$ by sticking together $E_{[f]}$ over one wedge-summand with $E_{[g]}$ over the other. In other words, $E'$ is made by clutching along an $S^3 \vee S^3$ by $f \vee g$. The pre-image of that $S^3 \vee S^3$ is an equator in the original $S^4$, and $f \vee g$ pulls back to a representative of $[f]+[g]$, and therefore $\pi^* E' = E_{[f]+[g]}$. That implies $p_1(E_{[f]}) + p_1(E_{[g]}) = p_1(E_{[f]+[g]})$.

As for $[f_{i+i',j+j'}] = [f_{ij}] + [f_{i'j'}] \in \pi_3(SO(4))$, first note that the map $h_i : S^3 \to S^3, u \mapsto u^i$ has degree $i$ (every point other than $\pm 1$ has $i$ pre-images). Since elements of $\pi_n(S^n)$ are classified by their degree, it follows that $[h_i] + [h_j] = [h_{i+j}]$.

P.S. After posting this, I realise that essentially the same question was answered in "Characteristic classes of sphere bundles over spheres in terms of clutching functions"

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