The torus knot page on Wikipedia contains the following (p,q)-torus knot parametrization:\begin{aligned}x&=r\cos(p\phi )\\y&=r\sin(p\phi )\\z&=-\sin(q\phi )\end{aligned} where\begin{aligned} r=\cos(q\phi )+2\end{aligned} and \begin{aligned} 0<\phi <2\pi\end{aligned}

What motivates the use of the additive inverse of the computed sine function?


The standard torus that a torus knot lies on has a coordinate system given by a longitude $\lambda$ and a meridian $\mu$.

Torus with a longitude and a meridian

The standard torus bounds a solid torus, which has a core circle. Give it an orientation, for example so that it has linking number $+1$ with the oriented $z$-axis. The longitude is a copy of the core circle pushed off to the boundary torus in such a way that the loop bounds a disk in the exterior (capturing the idea that $\lambda$ does not "wrap around" the torus about the minor circumference). The meridian is an oriented loop in the torus that has linking number $+1$ with the core circle and that bounds a disk in the interior of the solid torus (capturing a similar idea of not "wrapping around" longitudinally).

If you look at $\mu$ in the $XZ$ plane, you'll see that it is a circle that is oriented clockwise. That is why there is a minus sign.

Meridian in the XZ plane

I found it helpful to think about \begin{align*} x&= r = \cos(q\phi)+2\\ y&= 0\\ z&=-\sin(q\phi) \end{align*} which is the parameterization while being rotated about the $Z$ axis to keep the parameterized curve on the $XZ$ plane.

Homologically, a $(p,q)$ torus knot is $p[\lambda]+q[\mu]$. That is, it goes $p$ times around in the longitudinal direction while going $q$ times in the meridian direction.

  • $\begingroup$ Thank you @Kyle Miller for the visuals; they helped to illustrate your explanation. If the orientation of the red circle were counterclockwise, i.e. if there weren’t a minus sign, would the set of points the equations indicate no longer constitute a torus knot? If the points wouldn’t be a torus knot, which part of the definition of a torus knot would fail to apply to them? $\endgroup$ – bblohowiak Mar 31 at 19:56
  • $\begingroup$ @bblohowiak It would still be a torus knot (a torus knot is just a knot on a torus). But, it would be the $(p,-q)$ torus knot, which is different from the $(p,q)$ torus knot in general. $\endgroup$ – Kyle Miller Mar 31 at 19:59

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