Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
The answer will probably involve h-principles and the like which I have just started learning about. I eagerly await a good reply below. –  j.c. Dec 10 '09 at 2:43
    
I think h-principle will give us an immersed Lagrangian torus close to the original one, while I want an isotopy of the ambient space, which should give an embedded Lagrangian torus. –  Ilya Grigoriev Dec 10 '09 at 3:17

2 Answers 2

up vote 10 down vote accepted

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.

share|improve this answer

But there are non-trivial torus knots in $\mathbb R^4$. The simplest examples are achieved by attaching a handle to a knotted $S^2$ in $\mathbb R^4$. How do we know they're knotted? Most of these examples have complements with non-abelian fundamental group. Artin's spinning construction allows you to make knotted spheres in $\mathbb R^4$ from knotted circles in $\mathbb R^3$ -- in particular you can arrange for both knot complements to have the same fundamental group.

Or did you mean to add additional qualifiers to your question?

share|improve this answer
    
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! –  Ilya Grigoriev Dec 10 '09 at 18:05
    
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. –  Ryan Budney Dec 10 '09 at 23:55
    
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. –  Ryan Budney Dec 10 '09 at 23:58
    
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. –  Ilya Grigoriev Dec 11 '09 at 1:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.