Let $T(p,q)$ be a torus-knot where $p$ and $q$ are coprime. I am asking about the well known fact Which says if $p$ or $q$ is one, then $T(p,q)$ is trivial. If one of $p$ or $q$ is one, then we obtain a knot with crossings each of which can be eliminated by Reidemeister move 1. So at the end the knot is trivial. But I am wondering if we can apply Reidemeister move 1 in a Torus. The torus knot is a circle embedded in a standard torus. In fact $T(1,p)$ is a closed curve with $p$ number of full twists. I am wondering if we can apply Reidemeister move 1 to untwist the knot which is embedded in a torus. If it is embedded in 3-space, I agree. But I can not convinced my self how to deform it to a trivial such that all the deformations are done on a torus.
2 Answers
Any $p,q$ curve on the torus can be thought of as a representative of the homology class $(p,q) \in \mathbb{Z} \oplus \mathbb{Z} \approx H_1(\text{torus})$, so no, it cannot be deformed to a trivial knot within the torus.
As Lee states, this cannot be done on the torus and keep the torus in its standard embedding (obtain a knot which is trivial in the sense of crossings). But perhaps what you are asking is if we can deform the torus along with the knot under the Reidemeister moves, and the answer to this is, of course we can. As the three moves correspond to ambient isotopies (deforming space in a topologically preserving way), we can take the torus along for the ride of the unknotting. The only problem is that it will be hard to discern what the torus will look like, but it will still be a torus.
