$J(X)$ and exotic spheres.

Question

I read that we can realize exotic sphere as the coker of $J$-homomorphism. SO we can consider the exotic sphere $S^7$ realized using an identificantion of $B^4 \times S^3$ (where I donote the four dimentional ball). How can I realize this exotic sphere as coker of $J$-homomorphism? In particular, can I realize exotic spheres as coker of complex $J$-homomorphism $$ J:\pi_r(U(n)) \to \pi_{r+q}(S^q) ?$$

Answer 1

I don't know where you read this, but it is not quite true. What is true is that we have an exact sequence $$0 \longrightarrow \Theta_{n}/bP_{n+1} \longrightarrow \mathrm{coker}(J_n) \stackrel{K_n}{\longrightarrow} C_n \longrightarrow 0.$$ Here we have that

• $\Theta_n$ is the group of homotopy $n$-spheres.
• $bP_{n+1}$ is the subgroup of $\Theta_n$ consisting of homotopy $n$-spheres that bound a parallelizable $(n+1)$-manifold.
• $K_n$ is the Kervaire invariant.
• $C_n$ is a group that depends on $n$: $$C_n = \begin{cases} 0, & n \not\equiv 2 \pmod 4, \\ \Bbb Z/2 \,\text{ or }\, 0, & n \equiv 2 \pmod 4. \end{cases}$$

We have that $C_n = \Bbb Z/2$ for precisely those $n$ for which there exists an $n$-manifold of Kervaire invariant $1$. Finding which $n$ satisfy this property was a huge open problem in algebraic topology which was settled in 2009 for all cases except $n = 126$. It turns out that $C_n = \Bbb Z/2$ for $n = 6$, $14$, $30$, $62$, and possibly $n = 126$. Hence we have the following:

Theorem. For $n = 6$, $14$, $30$, or $62$, we have an exact sequence $$0 \longrightarrow \Theta_{n}/bP_{n+1} \longrightarrow \mathrm{coker}(J_n) \stackrel{K_n}{\longrightarrow} \Bbb Z/2 \longrightarrow 0. \tag{$\ast$}$$ For $n \neq 6$, $14$, $30$, $62$, or $126$, we have an isomorphism $$\Theta_n/bP_{n+1} \cong \mathrm{coker}(J_n). \tag{$\ast\ast$}$$ For $n = 126$, one of $(\ast)$ or $(\ast\ast)$ holds.