Knot Theory 3: Simplify, Simplify!

[stoutfellow]

The Story So Far: we've defined knots and links, and identified what it means for two knots to be equivalent. To study them more carefully, we've begun looking at plane representations of knots, and seen what that notion of equivalence translates to for plane representations: manipulation of the arcs between crossing points ("plane isotopy") and certain manipulations of the crossing points themselves ("Reidemeister moves").

Here's the problem. if two knots (or links) are equivalent, it's possible to manipulate any given plane representation of one, as just mentioned, so that it looks just like any given plane representation of the other. This is great, if you can come up with such a manipulation, but what if you can't? The best you can say, in that case, is that you don't know whether the knots are equivalent. We need a different, complementary approach, that of invariants. An invariant is a way of assigning, to any given knot or link, something - usually an algebraic object - in such a way that if two knots are equivalent the invariant assigns the same object to both. If you have an invariant, you can apply it to each of two knots; if the results are different, that tells you that the knots are not equivalent. (See the difference? The manipulation test is a way of showing that two knots are equivalent; invariants provide a way of showing that they aren't. The best of all situations is to have a complete set of invariants - one or more invariants with the property that, if each of them gives the same results to Knot1 and Knot2, then Knot1 and Knot2 are equivalent. Unfortunately, we don't have a complete set of invariants for knots as yet.)

To verify that a proposed invariant actually is an invariant, all we have to do is show that it isn't changed by plane isotopy or by Reidemeister moves. Since plane isotopy only affects the arcs in between crossing points, we can save time by looking at possible invariants which only look at the crossing points; then we only have to worry about the Reidemeister moves.

There are a number of known invariants for knots, but I'm only going to talk about one of them, the Jones polynomial. The invariants I've talked about in previous posts have been numbers, but this one is a more complex algebraic expression called a "Laurent polynomial". It is defined in terms of another Laurent polynomial, the "Kauffman bracket" - which is not an invariant, but can be tweaked to become one. The general strategy is this. First, we pick a value for the bracket of the simplest of all knots - the "unknot". Then, we develop a method of relating the bracket of any link to the brackets of simpler links - that is, if we know the brackets of those simpler links, we can compute the bracket of the original, more complex link. If we don't know those polynomials, we repeat the process. Each time, we get simpler and simpler links; eventually, we get down to a set of disconnected unknots - a bunch of circles, none of them twisted and none of them hooked together.

So, here are the rules. (Normally, the bracket is represented using angle-brackets, but I'm going to use square brackets instead, so as not to confuse LJ.)
1) [O] = 1. That is, the bracket of an unknot is just 1.
2) [L O] = -(A^2+1/A^2)[L]. In other words, if a link includes a disconnected unknot, throw it away; find the bracket of what's left, and multiply it by -(A^2 + 1/A^2). (Don't worry about what "A" is; it's just a variable, with no fixed value.) So, for example, consider the link consisting of two disconnected unknots, O O. [O O] = -(A^2 + 1/A^2) [O] = -(A^2 + 1/A^2).
3) If L is any link, pick any crossing point. We fiddle with the crossing point to produce two simpler links, which I'll call L1 and L2. Then [L] = A [L1] + 1/A [L2].

How do we construct L1 and L2? Turn your chosen crossing point so that the two strands pass (upper left - lower right) and (lower left - upper right):
X..Y
.\/
./\
Y..X
and so that the strand marked Y is on top. Now, cut the two strands; there are two ways to reconnect them:
X1...Y1
.\../
./..\
X1...Y1
(this gives us L1) and
X2.X2
.\/

./\
Y2.Y2
(which gives us L2).

Let's take an example. Take a circle and twist it into a figure-8; this is our L:
/\
\/
/\
\/.

This is really just an unknot. At the crossing point, the strand coming in from the upper right is on top. If we cut and reconnect the first way, we get L1:
/--\
\../
/..\
\__/.
(Raise the hyphens so they connect the tops of the slashes.) This is a simple loop, and so [L1]=1. If we cut and reconnect the second way, we get L2:
/\
\/

/\
\/.
This is two disconnected loops, and we saw before that [L2] = -(A^2 + 1/A^2). Therefore, [L] = A [L1] + 1/A [L2] = A + 1/A * -(A^2+1/A^2) = -1/A^3. Since the figure-8 was really just an unknot, we can see that the bracket isn't an invariant; if it were, our calculation would have come out to [O]=1.

It turns out, though, that we can correct the bracket to make it an invariant. To do this, we need to compute the "writhe" of the link, w(L). To do this, assign a direction to the link; for instance, with our figure-8, you might start at the top, move down to the right, then down to the left two steps, down to the right again, then up to the right, then up to the left twice, and finally up to the right; that brings us to our starting point again. Now, at each crossing point, look at the direction of the upper strand (in this case, that's the strand from the upper right, and it points down to the left), and the direction of the lower strand (down to the right). If you get from the upper direction to the lower direction by turning counterclockwise, write down a +1; if you have to turn clockwise, write down a -1. Do that for each crossing point, and add up the results; that's the writhe. In this case, we only have one crossing point, and it gives us a -1. So, the writhe of our figure-eight is -1. Now, take the writhe and multiply it by -3. Take A to that power, and multiply by -1. The Jones polynomial X(L) is the bracket, times the quantity we just computed. In our example, we get X(L) = -(A^-3)(-1/A^3) = 1, which is what we wanted.

So there's the strategy. Start with your link L. First compute the writhe of L. Then, compute the bracket of L. If you don't know the bracket of L, pick a crossing point and construct the simpler links L1 and L2. (They're simpler, because they each have one less crossing point.) If you know their brackets, you're done; if not, repeat the process. Sooner or later, you'll get to links which have no crossing points - that is, a bunch of disconnected unknots. Rules 1 and 2 allow us to compute their brackets, and then we can work our way back up to the original link L. Then combine the writhe and the bracket, as above, to get the Jones polynomial.

Now, I'll admit that this can be a tedious process, but it's a straightforward one, and it gives a definite answer; if two links have different Jones polynomials, they aren't equivalent. (If they do have the same Jones polynomials, then, well, you'll have to try something else; this isn't enough to tell you that they are equivalent.)


There are several other known invariants, all computed in similar ways; I hope the description under the cut gives some idea of the flavor of this sort of investigation.

Comments

Re: Knot Theory

[ndrosen.livejournal.com]

Thanks for the lessons in knot theory. I didn't follow it all, but it seems to be a good start for someone who can take the time to learn the stuff. I had just skipped over it until today, when I used the federal holiday to go back and read your various LJ postings on knots.