Consider the J-homomorphism $\pi_{n-1}(SO(n))\to \pi_{2n-1}(S^n)$ and $\tau_{S^n}:S^{n-1}\to SO(n)$ to be the adjoint of the classifying map of the tangent bundle of the standard sphere. In this paper, Wall claims that $J(\tau_{S^n})=[id_n,id_n]$ where $id_n$ is the identity map of $S^n$. How can one see this?
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1$\begingroup$ In principle I suppose you could also make a direct argument using framed cobordism. The pre-image of a point under the map $[id_n,id_n]$ is a standard linking pair $S^{n-1} \sqcup S^{n-1} \subset S^{2n-1}$, each trivially framed. The classifying map for the tangent bundle of $S^n$ can be described explicitly as a composite of two mirror reflections (with one map varying). Put those two together with the Hopf construction and that should suffice. $\endgroup$– Ryan BudneyCommented Jun 13, 2023 at 7:42
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1$\begingroup$ The relation follows from the discussion in Section 2 of James' "On the iterated suspension''. $\endgroup$– archipelagoCommented Jun 13, 2023 at 10:28
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There is probably more than one good way to look at this. I'll point out that $\tau_{S^n}$ is the boundary of a generator of $$ \pi_n(SO(n+1),SO(n))\cong \pi_n S^n. $$ The map $$ \pi_jS^n\cong \pi_j(SO(n+1),SO(n))\to \pi_{j+n}(\Omega S^{n+1},S^n) $$ is an isomorphism, not just for $j=n$ but for $j<2n-1$. And the attaching map for the $2n$-cell in the James model for $\Omega S^{n+1}$ is that Whitehead product.
answered Jun 12, 2023 at 12:51
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$\begingroup$ Great answer. Just a question: is there a reference for the connectivity of the map that you mentioned? $\endgroup$ Commented Jun 12, 2023 at 20:03
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2$\begingroup$ You can get it from the James model and the Blakers-Massey triad connectivity theorem (a.k.a. homotopy excision), and I suppose that someone did so about 75 years ago, but I don't have a reference. You can also prove it using functor calculus (maybe Weiss's orthogonal calculus). There is also an indirect proof using surgery theory! $\endgroup$ Commented Jun 12, 2023 at 20:51
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2$\begingroup$ The accepted answer of This question gives an argument for the connectivity of this map. $\endgroup$ Commented Jun 13, 2023 at 10:01