The question is pretty self-explanatory; we are dealing with the standard symplectic structure on ℝ4.


Some background: I'm reading the thesis "Lagrangian Unknottedness of Tori in Certain Symplectic 4-manifolds" by Alexander Ivrii, which proves that all embedded Lagrangian tori in ℝ4 are smoothly isotopic (and, in fact, Lagrangian isotopic). It uses lots of pseudoholomorphic curves. Obviously, if the question of this post is answered, together with the paper it will imply that all embedded tori in ℝ4 are smoothly isotopic (in other words, there are no torus knots in ℝ4).


I am told that this is, in fact, true, but that every proof that is known uses symplectic topology and Lagrangian tori. However, I have no idea how to do the question from the title, whether it's easy or hard, or whether it involves any pseudoholomorphic curves.



Why (and whether) is any smooth embedded torus in R^4 isotopic to an embedded Lagrangian torus?
Whoever told you that any embedded torus in R4 is isotopic to a Lagrangian torus was sorely mistaken. Luttinger (JDG 1995) observed the following: The manifolds obtained by doing certain Dehn-type surgeries on a Lagrangian torus in R4 admit symplectic structures. But the result X of the surgery is then always minimal and has the complement of a compact set symplectomorphic to the complement of a compact set in R4, whence a result of Gromov shows that X is actually symplectomorphic to R4. In particular X is simply connected. But for many knot types of tori in R4 (such as those mentioned by Ryan Budney) the results of these surgeries would in some cases not be simply connected, leading to the conclusion that these knot types must not admit Lagrangian representatives.


