Suppose one has a knot $K$ embedded in $\mathbb{R}^3$; but view $\mathbb{R}^3$ as a 3-flat in $\mathbb{R}^4$. Of course $K$ is not a knot in $\mathbb{R}^4$. I am wondering if there has been any study of how many "moves" are needed to unravel $K$ using the 4th dimension?

One might make this a sharper question in several ways. Here is one attempt. Say $K$ is represented as a 3D polygon of $n$ vertices in $\mathbb{R}^3$. A move consists of rotating a subchain of $C \subset K$ with endpoints $(a,b)$ into 4D and then back again at some new orientation into the 3-flat containing $K$. The endpoints $a$ and $b$ remain fixed, while $C$ is replaced by $C'$. Call the result knot $K'$.

Q1. Can every knot $K \subset \mathbb{R}^3$ be unraveled by these moves? If so...

Q2. How many such moves are needed to untangle $K$, as a function of some measure of $K$'s complexity, say, its crossing number?

These are entirely naïve questions, and I am not sure I have formulated them coherently. If not, apologies for the distraction!

Suppose one has a knot $K$ embedded in $\mathbb{R}^3$; but view $\mathbb{R}^3$ as a 3-flat in $\mathbb{R}^4$. Of course $K$ is not a knot in $\mathbb{R}^4$. I am wondering if there has been any study of how many "moves" are needed to unravel $K$ using the 4th dimension?

One might make this a sharper question in several ways. Here is one attempt. Say $K$ is represented as a 3D polygon of $n$ vertices in $\mathbb{R}^3$. A move consists of rotating a subchain of $C \subset K$ with endpoints $(a,b)$ into 4D and then back again at some new orientation into the 3-flat containing $K$. The endpoints $a$ and $b$ remain fixed, while $C$ is replaced by $C'$. Call the result knot $K'$.

Q1. Can every knot $K \subset \mathbb{R}^3$ be unraveled by these moves?

Answer: Yes, by Ian Agol's convincing argument.

Q2. How many such moves are needed to untangle $K$, as a function of some measure of $K$'s complexity, say, its crossing number?

These are entirely naïve questions, and I am not sure I have formulated them coherently. If not, apologies for the distraction!