Foliation of the 3-torus

[Anko]

TL;DR Summary

torus knots

I'm trying to understand more about the Hopf map and, I think I can see that the torus knot K1 defines the boundary of a looped, twisted ribbon embedded in the interior, aka the Mobius strip.

So slicing the torus open along the knot boundary means you have two halves of the torus linked together, so a K1 knot is a leaf. Is that correct or should I try something else?

 

[WWGD]

Maybe @mathwonk can help here?

 

[mathwonk]

pardon me I am not too familiar with foliations. what is a torus K1 knot? what is a leaf? for that matter what is the definition of a foliation? (presumably an equivalence relation whose classes are manifolds and with local product structure.)

what does the 3-torus have to do with the hopf map, (which is a circle bundle map from the 3 sphere to the 2 sphere)? can you provide some context? and what is a 3-torus? is it a solid torus, or is it a product of 3 circles? maybe @lavinia is the best person to ask.

One reason I am puzzled is you speak about cutting open the presumably 3 -dimensional "3-torus" apparently along a curve. That does not make sense to me. Of course you can cut open a 2-dimensional torus along a (k,1) knot, but you will get a disc, it seems to me, not a mobius strip. I could be wrong.

so we need to know the exact definition of the object you want to foliate, including its dimension. Then, if it is a 3-fold, (since you can presumably foliate a 3-fold with either 1 or 2 dimensional "leaves"), we need to know which one you want. then we need the exact construction of the leaves.

And if you spell all this out clearly, you will probably have answered your own question.

 

[Anko]

One reason I am puzzled is you speak about cutting open the presumably 3 -dimensional "3-torus" apparently along a curve. That does not make sense to me. Of course you can cut open a 2-dimensional torus along a (k,1) knot, but you will get a disc, it seems to me, not a mobius strip. I could be wrong.

I've seen drawings illustrating the embedding of a circle into the interior of a torus. The 3-torus as far as I understand it, is just the set of interior points of the 2-torus.

When you cut a torus open, you have to think about what the interior points are because now you want to deform the torus by flattening it out say, but the interior has to do something too, you can't just ignore it. What about contracting the torus so it gets thinner, etc?
What about cutting an actual doughnut open? how do you do that mathematically with a solid torus?

 

[jbergman]

I've seen drawings illustrating the embedding of a circle into the interior of a torus. The 3-torus as far as I understand it, is just the set of interior points of the 2-torus.

When you cut a torus open, you have to think about what the interior points are because now you want to deform the torus by flattening it out say, but the interior has to do something too, you can't just ignore it. What about contracting the torus so it gets thinner, etc?
What about cutting an actual doughnut open? how do you do that mathematically with a solid torus?

The a 1-torus is a circle, $S^1$. A 2 torus is a product of 2 circles, $S^1\times S^1$. The 3-torus is just, $S^1\times S^1 \times S^1$ and one could think of it as the generalization embedded in 4-d space.

 

[WWGD]

I've seen drawings illustrating the embedding of a circle into the interior of a torus. The 3-torus as far as I understand it, is just the set of interior points of the 2-torus.

When you cut a torus open, you have to think about what the interior points are because now you want to deform the torus by flattening it out say, but the interior has to do something too, you can't just ignore it. What about contracting the torus so it gets thinner, etc?
What about cutting an actual doughnut open? how do you do that mathematically with a solid torus?

Not sure of what you mean by how to do it. It can't be done continuously for one, and the resulting space will certainly not be homeomorphic to the torus. I assume you mean either the 1- or two tori here; $S^1$ or $S^1 \times S^1$.

 

[mathwonk]

the OP is apparently talking about the solid torus, S^1 x D, but as a subspace of the 3-sphere S^3. He is taking advantage of a "Heegaard" decomposition of S^3 into a union of two solid toruses, glued along their boundaries as manifolds, namely a common copy of S^1 x S^1. see this paper, esp. section 2.3 and Prop. 2.4, p.14, for this and the resulting relationship between links in S^3 and knots in S^1 x D.
https://arxiv.org/pdf/2008.03192.pdf

 

[Anko]

Does that paper answer my question, does the K₁
foliate the 2-torus and the 3-torus, after embedding the knot?

 

[mathwonk]

Apologies. Since I cannot understand why you are asking, and you have not clarified it, I just did some research and suggested a paper that may be relevant.