A polynomial vector field on $\mathbb{R}^3$ which has a knot periodic orbit

Question

Is there a polynomial vector field $$P(x,y,z)\partial_x+Q(x,y,z)\partial_y+R(x,y,z)\partial_z$$ which has a closed orbit $K$ such that $K$ is a non trivial knot?

Answer 1

For the Lorenz equations (which are polynomial), this was studied at least as far back as a 1983 paper in "Topology" by Birman and Williams -- MR0682059 if you have MathSciNet. The title is "Knotted periodic orbits in dynamical systems"; they prove, among other things, that the Lorenz flow has knots with arbitrarily high genus, and contains all torus knots.

There is also a 1997 volume of Lecture Notes in Mathematics entitled "Knots and Links in Three-Dimensional Flows", by R. Ghrist, P. Holmes, and M. Sullivan.

Both of these are referred to in the written version of the 2006 ICM address by Étienne Ghys, entitled "Knots and Dynamics", which is MR2334193 on MathSciNet if you have access, and is also posted on his website. There was a question on MSE some time back about a link to the video of his ICM talk, which may be of interest, and there is a nice website with many pictures by Ghys and Jos Leys that describes some of these ideas.

Answer 2

Here is an answer with many details to fill in. It will get you any knot.

Consider the closed solid torus $\mathbb D^2\times\mathbb S^1\subset\mathbb C^2$, and define on it the vector field $X:(x,u)\mapsto(-x,iu)$ (it goes along the torus, and is pushed to the center).

For any choice of (smooth) knot in any 3-dimensional manifold $M$, you can send the above torus to a neighborhood of it so that it winds in the same way. This induces a vector field on some neighborhood of your knot.

I claim that any field on $M$ whose restriction is close enough to the induced field will admit a closed orbit that flows once along the torus, hence is homotopic to the initial knot within the neighborhood. Indeed, provided the new vector field is still

• transverse to the discs $\mathbb D^2\times\{\theta\}$ and
• transverse to the boundary $\mathbb S^1\times\mathbb S^1$,

then there exists a well-defined Poincaré return map $f:\mathbb D^1\to\mathbb D^2$, it is continuous, and it sends the disc to a smaller disc. This means that, say by Brouwer's theorem, that $f$ has a fixed point, whose orbit must flow once along the torus by definition of the return map.

Now for your case, take any knot in $\mathbb R^3$, send the torus as described, and approximate the vector field to a high precision in this compact set by Stone–Weierstrass.

Answer 3

Over at math.SE, I give two polynomials $F(x,y,z)$, $G(x,y,z)$, whose common zero locus is the trefoil knot, and which are smooth and transverse there. Therefore, $(\nabla F) \times (\nabla G)$ is a vector field tangent to the knot.

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I think you should be able to get a shorter answer as follows. Take the sphere $S$ given by $x_1^2+y_1^2+x_2^2+y_2^2=1$ in $\mathbb{R}^4$. The vector field $$p(-y_1 \frac{\partial}{\partial x_1}+x_1 \frac{\partial}{\partial y_1})+q(-y_2 \frac{\partial}{\partial x_2}+x_2 \frac{\partial}{\partial y_2})$$ is tangent to $S$, and its flow lines are of the form $$(x_1, y_1, x_2, y_2) = (r_1 \cos(p t), r_1 \sin(p t), r_2 \cos(q t), r_2 \sin(q t))$$ which is a $(p,q)$ torus knot in $S$.

To give an even more extreme example, if we use $$(x_1^2+y_1^2)(-y_1 \frac{\partial}{\partial x_1}+x_1 \frac{\partial}{\partial y_1})+(x_2^2+y_2^2)(-y_2 \frac{\partial}{\partial x_2}+x_2 \frac{\partial}{\partial y_2})$$ then we should be able to get all the torus knots to occur in the same vector field: We'll have $T(p_1, p_2)$ on the torus $x_1^2+y_1^2 = \tfrac{p_1}{p_1+p_2}$, $x_2^2+y_2^2 = \tfrac{p_2}{p_1+p_2}$.

Now make a stereographic change of coordinates to get a vector field on $\mathbb{R}^3$ rather than $S^2$. That will give you a vector field with rational functions as coefficients, and clearing out the denominators will just change the speed of the flow.

But I don't feel up to doing that computation right now, so I'll leave the details to you.