6
$\begingroup$

Is it possible to have a one-dimensional foliation on three dimensional torus such that the foliation has a trefoil knot as its leaf? Moreover, does a one dimensional foliation on three dimensional torus have both compact and non-compact leaves?

Thank you very much!

$\endgroup$
11
$\begingroup$

[Apologies: the answer I wrote below is for $S^2 \times S^1$, not the 3-torus. Corrections added. The case of a 3-torus, or any 3-manifold, is included at the end.]

  1. First, there are different kinds of "trefoil knots" in $M = S^1 \times S^2$. Let's say we are interested in the "split trefoil knot", which lies in a 3-ball in $M$. This does in fact occur as a leaf of a foliation and it is easy to see.

  2. To build it, first recall the standard foliation of $S^3$ by circles where the one leaf is the trefoil knot $T$. This can be obtained by taking the orbits of the circle action $(z,w) \mapsto (t^2 z, t^3 w)$, $|t|=1$, on the 3-sphere in $\mathbb C^2$.

  3. Next we observe that the trefoil knot $T$ and a "disjoint" (split) unknot $U$ in $S^3$ can arise as the leaves of a foliation of $S^3$. (Split means there is an $S^2$ separating $U$ from $T$). To see this, consider the unknot $U'$ in (2) that arises as the orbit of $(z,w) = (1,0)$. This would almost do, except that $U'$ links $T$, i.e. $U'$ and $T$ are not split.

  4. To fix this, we use the fact (mentioned by Zare) that the space of constant slope foliations of a 2-torus is connected (it is just a circle). First adjust the foliation in (2) so it contains a solid torus $D^2 \times S^1$ foliated by $p\times S^1$'s with $U' = 0 \times S^1$. Here $D^2 = D^2(1)$ is the unit disk. By rotating the foliations on the tori $N_r = S^1(r) \times S^1$ as $r$ varies from 0 to 1, we can arrange that for some $r>0$ the torus $N_r$ is foliated by leaves of the form $U = S^1(r)\times p$, $p$ in $S^1$. Each of these leaves is an unknot split from $T$, so we have shown 3.

  5. To complete the picture, we now observe that $S^2 \times S^1$ can be obtained from $S^3$ by Dehn surgery along $U$, i.e. by cutting out a solid torus (disjoint from a ball containing $T$) with core curve $U$ and gluing it in again differently. Using the rotating foliation idea again, it is easy to extend the foliation from (3) to a foliation of $M$ with $T$ as a leaf. By construction, $T$ is a split trefoil in $M$.

Case of the 3-torus: It is also true that for any closed 3-manifold $M$ and any link $L$ in $M$, there exists a nowhere zero vector field on $M$ such that $L$ is invariant under the flow. One way to see this is to use Zare's construction, which depends on two facts: (a) $M$ can be obtained by Dehn surgery on a link $L'$ in $S^3$ (Lickorish) and (b) any link $L''$ in $S^3$ can be presented as a closed braid (Alexander). This construction will be a little less explicit than the one above, since it uses (a) and (b).

The main point is to prove the statement when $M=S^3$. For this one can start with any flow (e.g. the Hopf flow) such that there exists a closed trajectory. Thickening this flow line, we obtain an unknotted torus $S^1\times D^2$ with the product foliation. The foliation is obtained by suspending the identity map $D^2 \rightarrow D^2$. Now let us present $L$ as the closure of a braid $B$ with $n$ strands. Then $B$ can be obtained by suspending a diffeomorphism of $D^2$ to itself, fixing the boundary, and permuting $n$ points. Plug this new foliation into the original $D^2 \times S^1$. We then have a flow on $S^3$ such that $L$ is periodic.

For the general case, let $W = L' \cup L''$ such that $M$ is obtained by Dehn surgery on $L'$, and $L''$ becomes the desired link $L$ in $M$ after surgery. Then apply the construction above to $(S^3,W)$, and rotate the foliation near $L'$ as in step (5) above so it is compatible with Dehn surgery.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ $T^3$ cannot be obtained by Dehn surgery on the unknot in $S^3$, since $T^3$ has Heegaard genus 3, which is not a lens space. $\endgroup$ – Zhiyun Cheng Jul 30 '18 at 5:41
6
$\begingroup$

The $1$-dimensional foliations of $3$-manifolds are very flexible. You can have any tame knot $K$ as a closed leaf of a $1$-dimensional foliation of a $3$-torus.

Step 1: Foliate the torus by parallel circles, $S^1 \times \mathbb{T}^2$.

Step 2: Change the foliation near one circle $C_0$ to include a meridional circle $C_1$ with a product foliation around that. You can maintain cylindrical symmetry and that the leaves are tangent to tori around $C_0$. Change the angle with a longitude smoothly from $0$ to $\pi/2$, hold constant for an interval, then change the angle back to a multiple of $\pi$.

Step 3: Perturb the foliation near $C_1$ to include $K$ as a leaf with any representation of $K$ as a closed braid you want.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The knot leaf (trefoil knot) on 3-torus, I mean is that the bounded ball in 3-torus intersects with the leaf is the knotted 1-tangle. Hence, I think the braid you construct is not the one I wanted. Thank you! $\endgroup$ – Xifeng Su Aug 5 '16 at 6:29
  • $\begingroup$ What I constructed was a foliation with a trefoil knot as a leaf, which is what you asked. The trefoil is contained in a ball. If you mean that you want the knot to be nontrivial in homology, then I don't think it is normally called a trefoil knot, but it is easy to modify the construction to produce such a knot. $\endgroup$ – Douglas Zare Aug 6 '16 at 8:28
  • $\begingroup$ Would you please give some more details in step 2 and how is the knot K constructed in step 2? Some references and some pictures about the construction in step 2 are preferred. (I am working on dynamical systems and not quite familiar with your sentences of knot theory in step 2 and so I think I have some misunderstanding. I think that C_0 is a leaf, C_1 is another generator of the 2-torus other than C_0, the neighborhood you consider is the thickened torus C_0*C_1* I where I is the interval you mentioned above. Am I right? What is the cylindrical symmetry, the angle and "hold constant"?) $\endgroup$ – Xifeng Su Aug 7 '16 at 6:56
  • $\begingroup$ $C_0$ is a leaf of the foliation from step 1. It is not homologically trivial. $C_1$ is a meridional curve on a solid torus whose core is $C_0$. $C_1$ is homologically trivial. In step 2, I replace the foliation in a larger solid torus around $C_0$ so that it agrees with the old foliation on the boundary. Inside, all of the leaves are tangent to the tori of fixed distances from $C_0$. The induced foliations of the tori change in slope from the longitude to the meridian, $C_1$, then back to the longitude next to $C_0$. $\endgroup$ – Douglas Zare Aug 7 '16 at 12:05
1
$\begingroup$

We can foliate a torus (product of circles) by circle factors, with all leaves compact. If the torus has dimension 2, we can foliate it by lines with a given irrational slope quotiented down from the plane, and similarly in any dimension 2 or more, with no leaves compact. I don't know about knots.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Suppose that the three dimensional torus is the quotient of $R^3$ by the group $G$ generated by the translations $t_{e_i},=1,2,3$ of respective directions $e_i, i=1,2,3$. We denote by $p:R^3\rightarrow T^3$ the covering map.

The vector field $X$ of $R^3$ defined by $X_(x,y,z)=e_1+sin(2\pi y)e_2$ is invariant by $G$ and defines a vector field $Y$ on $T^3$, the orbit of $p(0,0,0)$ is compact, but the orbit through $p(0,1/8,0)$ is not compact.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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