braids and dynamics of roots of a polynomial

Question

The 2-variable polynomial equation $f(z,t) = 0$ with $z = \mathbb{C}, \\,t \in \mathbb{S^1}$ has $n = \mathrm{deg}_z f$ solutions each fixed $t$. I wanted to follow the roots as they travel with time paramter $t$ and count the number of distinct orbits. Is there a procedure for this?

Example: $f(z,t) = z^3 t^2 + z^2 t + z t^3 + z$ as our "time" parameter moves in a circle, the roots follow 2 distinct orbits. Here, $t \in \{ 0.9 e^{i \theta}: \theta \in [0, 2 \pi] \} = 0.9 S^1 \subset \mathbb{C}$

This defines a closed path in the configuration space of three points in the complex plane, i.e. a braid. My plot is a projection of this braid, forgetting the crossings. Looks like an circle + trefoil (not linked).

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EDIT: One point of the fundamental theorem of algebra is smooth deformations in the space of polynomials will not change the number of roots (if we projectivize). Proofs always involve deformations from arbitrary polynomials to $p(z) = z^n$ for some natural number degree $n = \deg p$.

If we add a periodic "time" variable, I expected each root to follow a disjoint closed path, but instead paths join together sometimes. So I started to ask about the number of components.

Geometrically, I would like to know if you can deform $\{ f(z,t)=0 \} \in \mathbb{C} \times S^1$ without changing the number of components. From the comments, it sounds like the number of components is constant unless the discriminant vanishes and whether it takes values in the unit circle. As long as these two events do not happen, is the topology of this set "constant"?

As an example: $f(z,t) = 1 + 3 z t + z^2 t + z^3 (2 + t)$ can be deform to $f_0(z,t) = z^3 + t z$ by "turning off" various coefficients. The number of components will be 2.

http://s7.postimage.org/dhvgpo1nd/polynomial_orbit.gif

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COMMENT: The closest thing I could find is the work of Vivek Shende and Alexei Oblomkov, where they study the intersection $\{ f(z,w) = 0 \} \cap \{ |z|^2 + |w|^2 = \epsilon\} \subset \mathbb{C}^2$ with $\epsilon << 1$ and $f(z,w)$ singular at $(z,w) = (0,0)$.

Topologically, it is a link in a 3-sphere. Generically it's trivial though it's possible to get torus knots for $f(z,w) = z^p + w^q$.

• arXiv:1003.1568 The Hilbert scheme of a plane curve singularity and the HOMFLY polynomial of its link. Alexei Oblomkov, Vivek Shende.

Comments in there point in various directions, e.g. Three-Dimensional Link Theory and Invariants of Plane Curve Singularities but it's on a different "setup".

Answer 1

Yes, the topology will be constant. The set of degree-$n$ polynomials with nonzero discriminant has fundamental group the braid group. You have a function from $S^1$ to this set, whose homotopy class defines an element of the braid group up to conjugation: a closed braid. This is of course the same closed braid you are measuring.

One probably doesn't need that machinery to prove that the closed braid is constant. You have a closed subset of the space $S^1 \times \mathbb C$ that is cut out by a certain equation, which is varying continuously. Because the equation never has zero derivative (because the discriminant never vanishes), changing the function a little will just move the closed set a little, which will not change the topology, or any topological invariant like the number of connected components.