# Cubics and Braids

## The bridge between the 3-strand braid group and the trefoil knot complement

Steve Trettel

| Topology, Algebra, Geometric Group Theory

Here’s a beautiful fact uniting algebra, topology, and group theory:

The n strand braid group may be described as the fundamental group of the configuration space of n (unordered) points in the plane. But this space is in a natural correspondence with the space of monic degree n polynomials over $\mathbb{C}$ with distinct roots - and so thought of as a topological space, this set of polynomials has fundamental group the n strand braid group!

In fact, in dimension 3 a particularly beautiful picture emerges: the space of generic cubic polynomials deformation retracts onto the trefoil knot complement, and this gives a particularly visual way to connect the trefoil knot group and the three strand braid group.

I want to take just a little time to explain this in this post here.

### Braid Groups

Given a positive integer $n$, the n-strand braid group $B_n$ is the algebraic structure encoding the different ways to braid together $n$ strands of string. The group operation here is concatenation, two braids are multiplied by stacking them on top of one another - and two braids are equivalent if they are isotopic to each other as collections of strings with their endpoints fixed on the top and bottom.

### Configuration spaces

A “configuration space of $n$ points in $M$”, or $C_n(M)$, is a topological space parameterizing the set of $n$ element subsets of $M$. As an example, let’s construct the configuration space of two points on a circle: a choice of two (ordered) points on a circle is just a choice of a point on $\mathbb{S}^1\times\mathbb{S}^1$, and requiring that they are distinct is the same as removing the diagonal $\Delta={(\theta,\theta)\mid\theta\in\mathbb{S}^1}$. To account for the fact that we want to parameterize collections of unordered points we must now quotient by the $\mathbb{Z}_2$ action swapping coordinates, and so

$$C_2(\mathbb{S}^1)\cong (\mathbb{S}^1\times\mathbb{S}^1\smallsetminus\Delta)/\mathbb{Z}_2$$

Its a fun exercise to convince yourself geometrically that this space is a mobius band!

The general construction is similar: we remove from the product $M^n$ the “big diagonal” $\Delta$, which is the set of points where at least two coordinates are equal and then quotient by the action of the symmetric group $\mathrm{SSym}_n$ by permuting coordinates. The case of interest to us here is actually rather simple, when $M$ is the plane, which we will view as $\mathbb{C}$. We then are just looking at the complement of the hyperplanes $z_i=z_j$ in $\mathbb{C}^n$ where we identify points whose coordinates are permutations of each other.

### The Fundamental Group and Braids

Let’s try to understand the fundamental groups of these spaces.
For a warm-up, what’s a loop in $C_2(\mathbb{C})$ look like? Well, as a point in this space is really a pair of distinct unordered points in $\mathbb{C}$, a loop in this space looks like a pair of points moving around $\mathbb{C}$ without colliding, and returning to themselves (setwise). In fact, fundamental the group $\mathbb{Z}$ generated by the path where two distinct points switch places!

This sort of reasoning generalizes: loops in $C_n(\mathbb{C})$ are just paths of $n$ distinct points in the plane moving about without bumping into each other and returning to themselves setwise.

A more instructive way to view this is to let the vertical direction be the “time axis” and watch the trails these points trace out as the move around and return to themselves: a path in $C_n(\mathbb{C})$ is just a tangle of strings and a loop is naturally a braid!

Thus the fundamental group of the $n$ point configuration space of $\mathbb{C}$ is none other than the $n$-strand braid group!
(Even better, its actually an Eilenberg-MacLane space for $B_n$).

### Spaces of Polynomials

(We already have one way of putting coordinates on the space $C_n(\mathbb{C})$: remember each point is described by a permutation class of elements of $\mathbb{C}^n$ with distinct coordinates. However it turns out theres an alternative (quite natural) way to do this using polynomials. A monic polynomial over $\mathbb{C}$ is completely determined by its roots, given roots $r_1,\ldots,r_n$ you can reconstruct the monic polynomial via $p=(z−r_1)(z−r_2)⋯(z−r_n$). And as the roots of a polynomial are naturally unordered, we may identify the $C_n(\mathbb{C})$ with the subset of $P_n(\mathbb{C})$ of polynomials with distinct roots!
Even better, this set is easy to describe: recall the discriminant $\Delta(p)$ of a polynomial is zero whenever there is a root of multiplicity greater than 1. This identification is unique up to scaling (if you scale a polynomial by a constant, it doesn’t change the roots), and so we should really work with the projectivization of the space of polynomials here. Doing all this, we find our configuration space is homemorphic to the projectivized complement of the discriminant locus:

$$C_n(\mathbb{C})\cong \mathrm{Proj}(P_n(\mathbb{C})\smallsetminus \Delta)$$

That means, theres a natural way to identify a loop in the space of polynomials with distinct roots (henceforth “generic” polynomials) with a braid - view the polynomials as their roots in $\mathbb{C}$ and then imagine the motion unfolding in three dimensions as a braid.

### Cubics and $B_3$

Let’s specialize for a moment to low dimensions, so that things are visualizable.
The case $n=1$ is trivial: the space of linear monic polynomials is naturally just $\mathbb{C}$. The case $n=2$ is slightly more interesting: $B_2≅\mathbb{Z}$ is generated by a braid with one “half twist” in two strands, and this is represented by the following path of quadratic polynomials for $t∈[0,\pi]$

$$p_t(z)=(z-e^{it})(z+e^{it})$$

The case n=3 is where things get really interesting: the space of generic cubics is now 6 dimensional (its an open subset of $\mathbb{C}^3$ when parameterized by the polynomial coefficients), and thats too high to see directly. Not to worry, we can cleverly perform a couple of deformation retractions which will retract the space of cubics onto a well-known three dimensional subspace! To start with, notice that when viewed as a collection of roots, each generic cubic is a (possibly degenerate) triangle in the plane, with the average of the roots the triangle’s center. For any given polynomial we may write down an intervals worth of translations of $\mathbb{C}$ starting at the identity and ending at the translation moving the center to 0, and its quick to verify that this process actually gives us a deformation retraction from the space of generic cubics to those with the average of their roots equal to 0 (the roots, and hence their average, are a continuous function of the coefficients….)

What has this done for us? Well, this homotopy has image a codimension 2 subset (we’ve essentially removed a planes worth of translations) and so we are dow to something that’s 4 dimensional! Even better - if we write our cubic as $p=z^3+az^2+bz+c$ its a standard fact that the coefficient a is the sum of the roots. So our we have actually landed on the subset of so-called “depressed cubics”: those of the form

$$p(z)=z^3+bz+c$$

Now the four dimensionality is clear: each such polynomial can be identified uniquely with the pair $(b,c)∈\mathbb{C}^2$. In this setting its even feasible to write down the discriminant explicitly:

$$\Delta(z^3+bz+c)=-4b^3-27c^2$$

Thus the space of generic cubics with root average zero is naturally identified with $\mathbb{C}^2∖{(b,c):4b^3=−27c^2}$. As this set clearly contains the point $(0,0)$ we might hope to be able to radially project the complement onto $\mathbb{S}^3$ to get a 3-dimensional picture, and this almost works. But instead of identifying all points $(tb,tc)$ along a ray in $\mathbb{C}^2$ we actually need to use ‘curved rays’ and say that points of the form $(t^2b,t^3c)$ are equivalent for $t>0$.

Lucky for us, this deformation retraction has geometric meaning - it is easy to verify that the roots of $p_t(z)=z^3+t^2bz+t^3c$ are simply a scaling of the roots of $z^3+bz+c$ by $t$ - and so what we have done is identify polynomials whose roots lie in scaling classes!

This is a space we have a good description of in coordinates: up to a linear change of variables our polynomials are coordinatized by

$${(b,c)\in\mathbb{C}^2\mid b^3\neq c^2, |b|^2+|c|^2=1}$$

Its a quick argument from here to see that this the trefoil knot complement! ( notice that $b^3=c^2$, $|b|^2+|c|^2=1$ forces both $b$ and $c$ to be of constant modulus, and so the solution lies on a torus in $\mathbb{S}^3$. But then $b^3=c^2$ forces this to be a $(3,2)$ cuve, which is a trefoil knot!)

Thus, given a cubic polynomial we have two distinct ways of viewing it: we may interpret its images under the appropriate homotopies as either its roots (a scaling class three distinct points in $\mathbb{C}$ with average $0$), or as its coefficients (a point in $\mathbb{S}^3$ off of the trefoil knot). Viewing a loop in the space of cubics through these two lenses shows us that it is both naturally a braid on three strands and a loop in the complement of the trefoil.

And so, the fundamental group of the trefoil complement is the three strand braid group.