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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).

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.

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 $f(z,w) = z^p + w^q$.

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".

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You'll have to be careful: sometimes, for some values of $t$, the roots of $f(z,t)$ might collide. That would result in you not quite getting a braid. (This is presumably obvious to you already, but it seems marginally worth stating). –  James Cranch Jun 27 '12 at 13:57
I'm not sure what you are asking. Barring intersections, you can follow the roots. Are you looking for a numerical method? –  Dylan Thurston Jun 27 '12 at 14:17
As long as the discriminant doesn't have any roots on the unit circle, you'll be fine. It seems to me that if you take a closed loop in the space of polynomials with non-zero discriminant, you should be able to take fourier approximants of the coefficients to get a polynomial in $S^1$ which closely approximates the path, and thus realizes any closed braid type. –  Ian Agol Jun 27 '12 at 14:42
Here's a couple of possibly relevant links: ams.org/mathscinet-getitem?mr=922801 homepages.math.uic.edu/~jan –  Ian Agol Jun 27 '12 at 14:56
is this descriminant = 0 or passing through $S^1$ the main obstruction? and I'd like to learn more about this Fourier decomposition... –  john mangual Jun 27 '12 at 15:24

1 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.

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