Skip to main content
6 events
when toggle format what by license comment
Mar 9, 2016 at 20:58 comment added Dylan Wilson The negative K groups are the easier part: $K^{-1}(\partial M)$ is just $K^0(\Sigma \partial M)$. The connecting homomorphism is just pulling back the bundle using the standard, geometric map $M/\partial M \rightarrow \Sigma \partial M$. (Build this by viewing $M/\partial M$ as collapsing a cone on $\partial M$ and just include into the `double cone' which is the suspension.)
Mar 9, 2016 at 18:50 history edited Stefan Kohl CC BY-SA 3.0
Fixed a typo.
Mar 9, 2016 at 18:30 review Close votes
Mar 13, 2016 at 22:28
Mar 9, 2016 at 18:23 comment added Denis Nardin Sort of. I suggest you look in Index theory for skew-adjoint Fredholm operators by Atiyah and Singer and Clifford modules by Atiyah, Bott and Shapiro for a description in terms of modules over a Clifford algebra.
Mar 9, 2016 at 17:38 review First posts
Mar 9, 2016 at 18:12
Mar 9, 2016 at 17:35 history asked student CC BY-SA 3.0