Low Dimensional Topology

February 15, 2011

Flowers on cocircularity

Filed under: Knot theory — dmoskovich @ 11:29 am

A couple of weeks ago, I went to Victoria to talk at a mini-workshop organized by Ryan Budney. A highlight for me was an interesting talk given by his graduate student, Garret Flowers, having to do with cocircularities on knots. He won “best graduate presentation” for a similar talk in UBC, which does not surprise me, because his work is tremendous fun, and it features the following two cult symbols in a mathematical context!

Check out his slides!

To obtain a knot diagram for a knot $K\colon\,S^1\hookrightarrow S^3$, imagine the knot sitting in the unit ball $B$. Place a light source up near $\{\infty\}$, and choose some plane $P$ which doesn’t interset $B$ to serve as a photographic plate. The knot strikes a pose (i.e. the embedding is fixed), the photographer chooses an angle (i.e. $P$ is fixed), the shutter snaps, and the photograph is taken.
A good photographer selects the best angle from which to snap a shot. The most popular angle is one in which the singularities in the knot diagram are as simple as possible- an angle which leads only to double-point singularities in the knot diagram- so that the diagram contains no self-tangencies or triple-points or funny stuff like that. A photojournalist would surely take such a photograph. But a fine art photographer might choose instead an angle which maximizes the number of arcs meeting at a point. And it turns out that, no matter the position of the knot, any non-trivial knot has a quadruple point from some angle (but generically, never a quintuple point). This, perhaps, is the angle which the fine art photographer will choose.
And so, we are led to define a quadrisecant as a line which intersects the knot at four points. Generically, a knot has only finitely many quadrisecants, as follows from a dimension count. Some of these are artifacts of the knot’s pose, while others are intrinsic to the knot’s topology. A photogenic knot would perhaps pose so as to minimize its quadrisecants, while a less photogenic knot might not do so. Which quadrisecants are essential?
Let’s switch over to thinking about long knots (smooth embeddings of $\mathbb{R}$ into $\mathbb{R}^3$ supported in the unit ball), which are the right setting for quadrisecants anyway. Orient everything. A quadrisecant $Q$ intersects a (long) knot $K$ at four points, which we’ll call $a$, $b$, $c$, and $d$.

There are two ways we could order these points:

1. As they appear as we walk along the long knot, from start to finish. We first encounter one point, then another, then another, and another. To fix things, let’s say we encounter $a$ then $b$ then $c$ then $d$.
2. As they appear along the quadrisecant, as we progress from one end of it to the other.

A quadrisecant is alternating if, along $Q$, we first see $c$, then $a$, then $d$, then $b$. Any non-trivial knot has an alternating quadrisecant, and these are the important ones. See Elizabeth Denne’s thesis.

Define the Gauss map to be the function which eats two points in $\mathbb{R}^3$, and spits out the direction from the first point to the second. Restricted to the knot, this becomes a map from $C_2(K)$ the configuration space of two distinct points on $K$ to the 2-sphere $S^2$. For an alternating quadrisecant, we want to consider the map from $C_4(K)$ to $(S^2)^3$ which maps an ordered quadruple of points $(x_1,x_2,x_3,x_4)$ on the knot to $\xi \stackrel{\textup{\tiny def}}{=} \left(\|x_3-x_1\|,\|x_1-x_4\|,\|x_4-x_2\|\right)$. Here’s the key point:
Fact: The intersection of $\xi$ with the diagonal $\Delta$ in $(S^2)^3$ is transversal.
This means that alternating quadrisecants can be used to construct long knot invariants. Indeed, Budney-Conant-Scanell-Sinha recovered the second coefficient of the Conway polynomial by summing quadrisecants with appropriate signs.
The key to the proof of transversality is that two points which are adjacent along the knot cannot be adjacent along the quadrisecant, and vice-versa. In fact, there’s an $\epsilon>0$, given by the tubular neighbourhood theorem, such that two points adjacent by the knot must be at least an $\epsilon$ apart along the quadrisecant. This would not be true for any other type of quadrisecant, hence alternating quadrisecants are indeed the important ones.

Satanic Circles

Taking the one-point compactification of $\mathbb{R}^3$ by adding a point $\{\infty\}$ a long knot becomes a knot, and a quadrisecant becomes a circle which intersects the knot and five points. So we obtain “cocircular” as opposed to “colinear” points along the knot. For a signed cocircularity count to give rise to a knot invariant, the image of an extended Gauss map on the knot must transversely intersect the space of cocircular sets of five points $x_1,x_2,x_3,x_4, x_5$ along the knot $K$. As in the case of quadrisecants, this can only happen if two points which are adjacent along the knot are not adjacent along the circle. Such cocircularities are suggestively called satanic circles.
Why satanic? Consider a circle $C$ which intersects $K\colon\,S^1\hookrightarrow S^3$ at five points $x_1,x_2,x_3,x_4, x_5$. These correspond to five points on the domain $S^1$ of $K$, which we draw as a circle. Connect each dot on $S^1$ to the dots you encounter immediately before and after it on $C$. Two points which are adjacent along the knot are not adjacent along the circle- the sign of the devil!
Counting satanic circles with appropriate signs recovers the second coefficient of the Conway polynomial, extending Budeny-Conant-Scanell-Sinha.

Watch the satanic circles dance as $\{\infty\}$ is moved around the knot!

Thelemic Circles

What about six point cocircularities? One may try the same thing. Define a thelemic circle to be a circle which intersects $K$ at six points. The naming is whim of mine for the purpose of this blog post- Garret calls them unicursal. To obtain an invariant from a signed count of thelemic circles, one again requires that the image of the extension of the Gauss map intersect transversely the space of cocircular sets of six points $x_1,x_2,x_3,x_4, x_5, x_6$ along the knot $K$. So again, two points adjacent along the knot cannot be adjacent along the circle. Connecting the dots on $S^1$ the domain of $K$ as before, we obtain another cult symbol- the unicursal hexagram, symbol of Thelema.
A signed count of these does give a knot invariant- unfortunately it’s trivial though. So Garret mods out by the action of the cyclic group of order 6 (a natural symmetry), which this time does give a knot invariant which is $\mathbb{Z}/2\mathbb{Z}$ values. Unfortunately it’s also trivial, which isn’t much fun, but it does give a non-trivial corollary: every knot has an even number of thelemic circles.