Calculate the linking number of two loops in $S^3$

Question

Given two loops $t\mapsto (sin(t),cos(t),sin(t+\phi),cos(t+\phi))$ and $t\mapsto (sin(t),cos(t),sin(t),cos(t))$ for $\phi\in]0,2\pi[$. How to show that their linking number is one?

I appreciate any help.

Answer 1

Linking number is a continuous function of $a$, and it is undefined whenever the two loops intersect. Notice that the two loops intersect only when they are the same loop, when $\phi=2\pi n$ with $n\in\mathbb{Z}$. Hence, on your domain for $\phi$, linking number is constant.

Now, let's pick a convenient value for $\phi$ to calculate the linking number. Probably right in the middle of the domain will work well: $\phi=\pi$.

Like I mentioned in a comment, the paths you have are not normalized to lie on the unit sphere, but they become normalized by dividing by $\sqrt{2}$.

It is not too hard to come up with an orthogonal matrix that carries the path $h(t)=(\cos(t),\sin(t),0,0)$ to the second path. Here is one: $$ A=\left( \begin{array}{cccc} 0 & \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} & 0 \\ \end{array} \right)$$ Then, $A(0,0,\cos(t),\sin(t))$ is the first path at $\phi=\pi$. Thus, the problem comes down to finding the linking number between $t\mapsto (\cos(t),\sin(t),0,0)$ and $t\mapsto (0,0,\cos(t),\sin(t))$.

I find it easiest to contemplate linking number in a stereographic projection. I will make the odd choice of projecting from $(0,0,0,1)$, odd since the second of the loops passes through it. Projected, the paths are $t\mapsto (\cos(t),\sin(t),0)$ and $t\mapsto (0,0,\frac{\cos(t)}{1-\sin(t)})$.

Notice that the first loop wraps around the $z$ axis counter-clockwise once, and the second loop is the $z$ axis. There is a single singularity at $t=\pi/2$, so this loop travels along the $z$ axis in the positive direction and passes through $\infty$ once. This is enough to know that the linking number is $+1$.