# Can I solve an integral (or other tough problem) by playing with knots?

I've seen that in calculating things in knot theory that involves a lot of hard looking integrals and matrices, even though the knots themselves appear fairly simple.

So is there some way in which this works backwards, where I might have a really large determinant or complicated integral that can be represented as a knot, and by tying, untying, or rearranging the knot I can solve the problem on that "knot domain" and then transfer it back into the lingo of integrals and matrices, kind of like you might imagine using a laplace transform?

• Great question. Jan 24, 2014 at 23:25
• Perhaps it would be possible to define any polynomial as the knot polynomial of some set of knots and then perform the normal algebraic operations?
– user122283
Jan 24, 2014 at 23:34
• @SanathDevalapurkar That sounds fun and simple, but I honestly know little to nothing about knot theory so I really came up with this question in trying to self-teach off of the internet. If you could point me to a good starting place, I think I've read about a billion times that knot theory was originally created by Lord Kelvin as an atomic theory, and on the other hand there are these fairly formidable looking integrals with an archaic explanation with little to no middle ground on the internet. Jan 25, 2014 at 1:59
• @Kainui: Sure! To give an introduction, I suggest matapp.unimib.it/~ricca/teaching/Torino1.pdf or maths.manchester.ac.uk/~kd/ma351/ma351.html.
– user122283
Jan 25, 2014 at 3:39

I like your question. Unfortunately, I do not have a general answer, but I do have a specific example where it is possible. You can look more into this if you wish in Rolfsen, Knots and Links under linking numbers, which is my reference. A link is just a few disjoint knots, which may be "linked" together. There is something called a Gauss Integral $\ell k (J,K)$ defined for a link which has components $J$ and $K$ which is
$$\frac{1}{4\pi} \int_J \int_K \frac{ (x'-x)(dxdz'-dzdy')+(y'-y)(dzdx'-dxdz')+(z'-z)(dxdy'-dydx') }{ [(x'-x)^2+(y'-y)^2+(z'-z)^2]^\frac{3}{2} }$$
where $(x,y,z)\in J$ and $(x',y',z')\in K.$ This is obviously an integral you do not want to have to work out. Luckily, Rolfsen gives seven equivalent ways of computing the linking number. The most common of which is the following: For an oriented link diagram of $J\cup K$, only look at the crossing where $K$ is the overarc (on top). Then rotate the picture until $K$ is "pointing" up at one of the crossing. If $J$ goes from right to left, assign a $+1$ to this crossing. If $J$ goes from left to right, a $-1$. The sum of these is your linking number, up to a sign convention. So, you never have to do the integral!