Timeline for Involutions on $D^4$ with a fixed arc
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 20, 2021 at 3:13 | comment | added | Strongly Negative Amphicheiral | Ah, thanks! Yes, that's my suspicion as well. | |
| Mar 20, 2021 at 3:06 | comment | added | Igor Belegradek | It has to be an exotic $D^2\times RP^2$ whose 2-fold cover is the standard $D^2\times S^2$. Then you can glue it back. But I suspect such exotic $D^2\times RP^2$'s are unknown. (I am not a 4d topologist of course). | |
| Mar 20, 2021 at 2:56 | comment | added | Strongly Negative Amphicheiral | Thanks for the comment! I like the idea of excising the fixed set, but I'm not sure I follow the last part. What if the double cover of the exotic $D^2 \times \mathbb{R}P^2$ is an exotic $D^2 \times S^2$? Then I don't immediately see how to get a smooth involution on the standard $D^2 \times S^2$ (or on the standard $D^4$). | |
| Mar 20, 2021 at 1:16 | comment | added | Igor Belegradek | For example, is there an exotic $D^2\times RP^2$? If yes, you can reverse the above construction, and get a non-standard involution on $D^4$ with fixed point set an arc. | |
| Mar 20, 2021 at 1:09 | comment | added | Igor Belegradek | The group action can be eliminated from the discussion by excising a tubular neighborhood of the fixed point set. Start from a given an involution of $D^4$, and attach along $\partial D^4$ the standard involution (=product of the identity map of $D^1$ and the antipodal map of $D^3$ with corners smoothed). You get a involution of $S^4$ that fixes an unknotted circle. Excising an equivariant tubular neighborhood of the circle, and passing to the quotient gives a 4-manifold with boundary $S^1\times RP^2$ whose 2-fold cover is $D^2\times S^2$. You need to smoothly classify such manifolds. | |
| Mar 19, 2021 at 23:57 | history | asked | Strongly Negative Amphicheiral | CC BY-SA 4.0 |