Timeline for Why (and whether) is any smooth embedded torus in R^4 isotopic to an embedded Lagrangian torus?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2009 at 1:39 | comment | added | Ilya Grigoriev | Thanks again, this is all very useful! I can't mark both answers as "correct", so I guess I'll pick the one with less reputation. | |
| Dec 10, 2009 at 23:58 | comment | added | Ryan Budney | I guess technically not all embeddings of $S^1 \times D^2$ in $\mathbb R^4$ are isotopic, but their boundary tori are. ie: there are precisely two isotopy classes of embeddings of $S^1 \times D^2$ in $\mathbb R^4$ and they differ by a full meridional twist. | |
| Dec 10, 2009 at 23:55 | comment | added | Ryan Budney | The fundamental group of the complement of an unknotted torus in $\mathbb R^4$ is the integers. To be concrete, let's consider a torus in $\mathbb R^4$ to be unknotted if it is the boundary of an embedded $S^1 \times D^2$ -- all embeddings of $S^1 \times D^2$ in $\mathbb R^4$ are isotopic. Rolfsen's book "knots and links" has a very basic treatment of spinning. A. Kawauchi's book "A survey of knot theory" has a more in-depth treatment. Spinning fits into a bigger context of homotopy long exact sequences for pseudo-isotopy embedding spaces, but that's another story. | |
| Dec 10, 2009 at 18:05 | comment | added | Ilya Grigoriev | Thank you very much for the quick answer! Unfortunately, I know almost nothing about knot theory. Do you have a good references for the spinning construction and for why nontrivial fund. gp. implies nontrivial knot? Thanks! | |
| Dec 10, 2009 at 3:40 | history | edited | Ryan Budney | CC BY-SA 2.5 |
added details; added 6 characters in body
|
| Dec 10, 2009 at 3:30 | history | answered | Ryan Budney | CC BY-SA 2.5 |