Skip to main content
8 events
when toggle format what by license comment
Feb 26, 2021 at 19:10 comment added archipelago The fact that the kernel of $\eta^B$ for $B=\tau_{\ge k+1}BO$ (this is $BO\langle k\rangle$ in your notation, if I am not mistaken) is trivial follows from classical surgery, Kervaire--Milnor style: If $\Sigma$ is in the kernel of $\eta^B$, then it bounds a $k$-parallelisable $(2k+1)$-manifold. Doing surgery on in the interior of $W$, we see that $\Sigma$ bounds a $k$-connected $(2k+1)$-manifold $W'$. By Poincaré duality, $W'$ is contractible, so $W'\cong D^{2k+1}$ by the $h$-cobordism theorem, and thus $\Sigma\cong S^{2k}$.
Feb 26, 2021 at 18:53 comment added archipelago Here $\overline{\Sigma}$ is $\Sigma$ with the opposite orientation---the inverse in the group of homotopy spheres.
Feb 26, 2021 at 18:51 comment added archipelago Just to clarify: no appeal to Kreck's work is necessary to answer the original question when exotic spheres are stably diffeomorphic. $\Sigma \sharp W_g\cong \Sigma'\sharp W_h$ implies $g=h$ by consulting the Euler characteristic. But then $W_g\cong \overline{\Sigma}\sharp \Sigma \sharp W_g\cong \overline{\Sigma}\sharp \Sigma'\sharp W_g$ and thus $\overline{\Sigma}\sharp \Sigma'\cong S^{2n}$ since the inertia group is trivial, so $\Sigma\cong\Sigma'$.
Feb 26, 2021 at 18:12 vote accept Chris Schommer-Pries
Feb 26, 2021 at 17:25 comment added Chris Schommer-Pries Thank you! Your references answer my question, but I wanted to clarify something. The inertia group being trivial means that $W_g$ and $W_g \# \Sigma$ are only diffeomorphic if $\Sigma = S^n$ is standard. By Kreck's result that means and exotic $\Sigma$ must be different in $BO\langle k \rangle$-bordism from the standard sphere. So $\eta^B$ is injective and the kernel is zero. Agreed?
Feb 26, 2021 at 16:43 history edited Jens Reinhold CC BY-SA 4.0
added 9 characters in body
Feb 26, 2021 at 16:37 history edited Jens Reinhold CC BY-SA 4.0
added 56 characters in body
Feb 26, 2021 at 16:20 history answered Jens Reinhold CC BY-SA 4.0